To take the longest possible jump, an athlete should make an angle of _____

Option 3 : 45 degree with the ground

**Concept:**

Projectile motion:

When a particle is projected obliquely near the earth's surface, it moves simultaneously in horizontal and vertical directions. The **path** of such a particle is called **projectile** and the motion is called **projectile motion**.

**Range of projectile:**

- The horizontal range of a projectile is the
**distance along the horizontal plane**it would travel, before reaching the same vertical position as it started from.

**Formulae in projectile motion:**

\(Range\;of\;projectile = \frac{{{u^2}\sin 2θ }}{g}\)

\(Total\;time\;of\;flight = \frac{{2u\;sinθ }}{g}\)

\(Maximum\;Height = \frac{{{u^2}{{\sin }^2}θ }}{{2g}}\)

where u = projected speed, θ = angle at which an object is thrown from the ground and g = acceleration due to gravity = 9.8 m/s^{2}.

**Calculation:**

**Given:**

Range of a Projectile motion is given by (R):

\(R= \frac{{{u^2}\sin 2θ }}{g}\)

For horizontal distance to be maximum:

sin 2θ = 1

∴ sin 2θ = sin 90°

∴ θ = 45°.

**∴ to take the longest possible jump, an athlete should make an angle of 45° with the ground.**

Option 2 : 30° with vertical

__Concept:__

For the maximum range in case of the inclined plane, the angle from the inclined should be,

\(θ = \frac{\pi}{4} - \frac{α}{2}\)

where, θ = angle of projection from the inclined plane and α is the angle of incline from horizontal.

__Calculation:__

__Given:__

The angle of incline, α = 30°

For maximum range,

\(θ = \frac{\pi}{4} - \frac{α}{2}\)

\(θ = 45 - \frac{30}{2}\)

θ = 30°.

Angle of projection from the vertical = 30°.

Option 3 : 6 s

__Concept:__

- It is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
- Initial Velocity: The initial velocity can be given as x components and y components.

- Component of initial velocity in x-direction, (ux) = u cos θ
- Component of initial velocity in the y-direction, (uy) = u sin θ
- In the case of projectile motion, we can see a free-fall motion of a body on a parabolic path with constant velocity.
- If a body is thrown at a certain angle then during its movement, we get two components of velocity as given below.

- And thus, the range of a projectile is the displacement of a particle along the x-axis and can be given as:

The **height of projectile is given by,**

**\(H_2~=~\frac{u^2_y}{2g}\)**

- Whereas the time of flight is the total time for which projectile stayed in the air.

Time of flight for the projectile,

\(\left( {\bf{t}} \right) = \;\frac{{2 ~u ~sin~θ }}{g}~=~\frac{2u_y}{g}\)

where The angle of projection = θ, Initial velocity = u, Gravitational acceleration = g, Time of flight = t, Range of projectile = R

__Calculation:__

**Given****:**

Height of the tower H_{1 }= 60 m, initial velocity of the object** u** = 40 m/s, angle of inclination with horizontal** θ** = 30°,** u _{x}** = 40 × cos 30° = 20√3 m/s,

Time taken to reach the maximum height is given by,

\(T_1~=~\frac{u_y}{g}~=~\frac{20}{10}~=~2~s\)

Height above the tower is given by,

\(H_2~=~\frac{u^2_y}{2g}~=~\frac{20~\times ~20}{2~\times~ 10}~=~20~m\)

Therefore, the maximum height is,

H_{max} = H_{1 }+ H_{2} = 60 + 20 = 80 m

Now, the time taken to free fall from maximum height is,

\(T_2~=~\sqrt{\frac{2H_{max}}{g}}~=~\sqrt{\frac{2~\times ~80}{10}}~=~4~s\)

Thus, the total time is taken during the entire flight is given by,

**T _{total} = T_{1 }+ T_{2} = 2 + 4 =**

Option 2 : \(\sqrt {\frac{{2H}}{g}} \)

__Concept:__

At the highest point of the projectile:

For projectile motion we know: a_{x }= 0 and a_{y }= -g

**Diagram**

**Explanation:**

V_{y }= 0 (at highest point)

\(V_y^2 - u_y^2 = - 2gH\)

u^{2 }sin ^{2}θ = 2gH

u sinθ = \(\sqrt {2{\rm{gH}}} \)

V_{y }= u_{y}+ a_{y}t

0 = \(\sqrt {2{\rm{gH}}} \) - gt

⇒ Time taken by the projectile to reach the highest point,

t = \(\sqrt {\frac{{2{\rm{H}}}}{{\rm{g}}}} \)

Option 3 : \(\frac{u^2 sin^2\theta}{2g}\)

__Concept:__

Projectile motion:

Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.

Initial Velocity: The initial velocity can be given as x components and y components.

ux = u cosθ

uy = u sinθ

Where u stands for initial velocity magnitude and θ refers to projectile angle.

Maximum height: It is the maximum height from the point of projection, a projectile can reach

The mathematical expression of the maximum height is -

\(\Rightarrow H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

__Important Points__

Time is taken to reach maximum height:

it is half of the total time of flight.

\(\Rightarrow {{\rm{T}}_{1/2}} = \frac{{{\rm{v\;sin\theta }}}}{{\rm{g}}}\)

Where T1/2 = time taken by the projectile to reach maximum height, g = acceleration due to gravity, and v = velocity

Time of Flight: The time of flight of projectile motion is the time from when the object is projected to the time it reaches the surface.

\(\Rightarrow {\rm{T}} = \frac{{2{\rm{\;v\;sin\theta }}}}{{\rm{g}}}\)

Where T is the total time taken by the projectile, g is the acceleration due to gravity.

Range: The range of the motion is fixed by the condition y = 0.

\(\Rightarrow R = \frac{{{v^2}sin2\theta }}{g}\)

Where R is the total distance covered by the projectile.

Option 1 : \(\frac{{2{\rm{u}}\sin \theta }}{{\rm{g}}}\)

__Explanation:__

Projectile motion: A kind of motion that is experienced by an object when it is projected near the Earth's surface and it moves along a curved path under the action of gravitational force.

- When a particle moves in projectile motion, its velocity has two components.
- vertical component (u sinθ)
- horizontal component (u cosθ)

- There is no acceleration or forces in the horizontal direction so the horizontal component of velocity (u cosθ) will remain constant.

The range of the projectile,

R = u cos θ × t

Whereas the time of flight is the total time for which the projectile stayed in the air.

Time of Flight:

The time of flight of projectile motion is the time from when the object is projected to the time it reaches the surface.

- The mathematical expression of the time of flight is -
- \(\frac{{2{\rm{u}}\sin \alpha }}{{\rm{g}}}\)

The angle of projection = θ, Initial velocity = u, Gravitational acceleration = g, Time of flight = t, Range of projectile = R

Maximum height:

It is the maximum height from the point of projection, a projectile can reach

- The mathematical expression of the
**maximum height**is - - \({\rm{H}} = \frac{{{{\rm{u}}^2}{{\sin }^2}\theta }}{{2{\rm{g}}}}\)

**Horizontal Range:**

The range of the motion is fixed by the condition y = 0.

- The mathematical expression of the horizontal range is -
- \({\rm{R}} = \frac{{{{\rm{u}}^2}\sin 2\theta }}{{\rm{g}}}\)

Where R is the total distance covered by the projectile.

Option 1 :

3 v cos θ

**Concept:**

Initial momentum \( = m\left[ {v\cos \theta \hat i + v\sin \theta \hat j} \right]\)

Final momentum \( = - \frac{m}{2}v\cos \theta \hat i + \frac{m}{2}v'\hat i\)

Hence conserving momentum

\(mv\cos \theta = - \frac{{mv}}{2}\cos \theta + \frac{m}{2}v'\)

\(\frac{{3m}}{2}v\cos \theta = \frac{m}{2}v'\)

∴ v’ = 3 v cos θ

Option 3 : 45°

__Explanation__:

Oblique Projectile Motion:

- In projectile motion, the horizontal component of velocity (u cosθ ), acceleration (g), and mechanical energy remain a constant while, speed, velocity, the vertical component of velocity (u sinθ), momentum, kinetic energy and potential energy all changes.
- Velocity and KE are maximum at the point of projection is minimum (but not zero) at the highest point.

Horizontal range –

- It is the horizontal distance traveled by a body during the time of flight.

- The mathematical expression of the horizontal range is

\(\Rightarrow R=\frac{{{u}^{2}}\sin 2\theta }{g}\)

- As we know that sin θ is maximum at 90°.
- Therefore horizontal range will maximum at 45°.

\(\Rightarrow R=\frac{{{u}^{2}}\sin 2\times 45{}^\circ }{g}=\frac{{{u}^{2}}\sin 90{}^\circ }{g}=\frac{{{u}^{2}}}{g}\)

- When θ = 90°, then the horizontal range will be

\(\Rightarrow R=\frac{{{u}^{2}}\sin 2\times 90{}^\circ }{g}=\frac{{{u}^{2}}\sin 180{}^\circ }{g}=\frac{{{u}^{2}}\sin \left( 90{}^\circ +90{}^\circ \right)}{g}=\frac{{{u}^{2}}\cos 90{}^\circ }{g}=0\) [As, cos 90° = 0]

- The horizontal range will be minimum if the angle of projection is 90°.

Option 2 : \(h=\frac{u^2 \sin^2 \theta}{2g}\)

__Explanation:__

Projectile motion: Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.

The maximum height is attained at point A and using the kinematic equation it can be given as:

\({v^2} = {u^2} + 2as\)

where

**s **= maximum height **H**

**v** = Final velocity in the y-direction which is zero for maximum height = 0

**u** = Component of initial velocity in the y-direction **uy = usinθ**

**g **= acceleration due to gravity the body will be decelerated due to gravity hence we took it as negative

\({v^2} = {u^2} + 2as \Rightarrow 0 = u_y^2 - 2gH\)

\(H = \frac{{u_y^2}}{{2g}} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

__Important Points__

- For a projectile, the
**maximum height attain will depend on the Y component of its initial velocity (uy)**. - we have to consider the acceleration due to gravity because as a particle is thrown upward due to gravitational acceleration its velocity will keep on decreasing and at
**extreme position uy becomes zero**and body accelerated downward hence it starts falling again - we took acceleration an as acceleration due to gravity g, whereas the
**body will be decelerated due to gravity hence we took it as negative.** **Initial Velocity:**The initial velocity can be given as x components and y components.- Component of initial velocity in x-direction, (ux) = ucosθ
- Component of initial velocity in the y-direction, (uy) = usinθ

- Time of flight, \(t = \frac{{2{u}\sin \theta }}{g}\)
- Horizontal range, \(R = \frac{{u ^2\sin 2\theta }}{g}\)

Option 1 : Parabolic

**Projectile motion:**

If a particle is projected in the air with some oblique angle then the particle traces a path and falls on the ground at some point. The particle is called as projectile and its motion in the air is called projectile motion.

**The equation of trajectory for the projectile is given by**

\({\rm{Y}} = {\rm{X}}\tan {\rm{\alpha }} - \frac{{{\rm{g}}{{\rm{X}}^2}}}{{2{{\rm{u}}^2}{{\cos }^2}{\rm{\alpha }}}}\)

Where,

u = Velocity of projection

α = Angle of projection

The equation is in the form Y = AX + BX^{2}

Where Y = AX + BX^{2} is the equation of parabola

Hence the path traced by a projectile is parabolic

__Important point:__

**Terms related to projectile motion**

**1)Time of flight:**

It is the duration of time for which projectile remains in the air

\({\rm{T}} = 2{\rm{u}}\sin \frac{{\rm{\alpha }}}{{\rm{g}}}\)

**2) Horizontal range:**

It is the horizontal distance between the point of projection and point of landing.

\({\rm{R}} = \frac{{{{\rm{u}}^2}\sin 2{\rm{\alpha }}}}{{\rm{g}}}\)

**3) Height:**

It is the maximum vertical distance travelled by projectile.

\({\rm{H}} = \frac{{{{\rm{u}}^2}{{\sin }^2}{\rm{\alpha }}}}{{2{\rm{g}}}}\)

Option 3 : 2 s

**Concept:**

**Time to reach max height = ****\(\frac {u}{g}\)**

Total time of flight = 2 × \(\frac {u}{g}\) = \(\frac {2u}{g}\)

**Calculation:**

**Given:**

u = 9.8 m/s

Time of flight = **\(\frac {u}{g}= \frac {9.8}{9.8}= 1 ~s\)**

Total time of flight = 2 × 1 = 2 s

Option 3 : A car moving in a straight line

**Explanation:**

**Projectile motion:**

- A kind of motion that is experienced by an object when it is projected near the Earth's surface and it moves along a curved path under the action of gravitational force.
- Projectile motion is the motion of an object thrown or projected into the
**air**, subject to**only the****acceleration****of gravity**. The object is called a projectile, and its path is called its trajectory. - In case of a stone thrown in any direction, a stone thrown horizontally from a building, and a bullet fired from a gun are subjected to only gravitional force. But in case of a car moving in a straight line is a rectilinear type of motion.

Option 2 : 45°

__Explanation:__

**projectile motion:**

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity.

**In Projectile motion:**

\(R = \frac {V_0 ^2 sin2θ }{g}\)

where R = Horizontal Range, V_{0} = Initial velocity, θ = The angle at which object is projected or thrown.

For maximum value of 'R' sin2θ should be maximum

Maximum value of Sin2θ = 1

∴ 2θ = 90°

∴ θ = 45°

**Hence, horizontal range will be maximum when object is projected at 45° angle with the horizontal.**

__Additional Information__

**Time of flight = \(\frac {2V_0 sin\theta }{g}\)**

**Time to reach maximum height** = **\(\frac {V_0 sin\theta }{g}\)**

**Maximum height = H _{max} =**

Option 1 : R = (u^{2}sin2θ) / g

__Concept:__

- Projectile motion: Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
- Initial Velocity: The initial velocity can be given as x components and y components.

- Component of initial velocity in x-direction, (ux) = ucosθ
- Component of initial velocity in the y-direction, (uy) = usinθ
- In the case of projectile motion, we can see a free-fall motion of a body on a parabolic path with constant velocity.
- If a body is thrown at a certain angle then during its movement, we get two components of velocity as given below.

- And thus, the range of a projectile is the displacement of a particle along the x-axis and can be given as:

The range of the projectile,

- Whereas the time of flight is the total time for which projectile stayed in the air.

Time of flight for the projectile, \(\left( R \right)\; = \;ucos\theta \times t\)

The angle of projection = θ

Initial velocity = u

Gravitational acceleration = g

Time of flight = t

Range of projectile = R

__Explanation:__

As given above,

v = initial velocity of projectile

θ = angle with x-axis

Time of flight for the projectile, \(\left( {\bf{t}} \right) = \;\frac{{2vsin\theta }}{g}\)

And the range of the projectile, \(\left( R \right)\; = \;vcos\theta \times t\)

Thus, by comparing the above two equation range of projectile can also be modified as

\(R = vcos\;\theta \times \frac{{2vsin\;\theta }}{g} \Rightarrow R = \frac{{2{v^2}sin\theta cos\theta }}{g}\)

But \(2sin\theta .cos\theta = sin2\theta\) (By using trigonometric relation)

\(\Rightarrow R = \frac{{{v^2}sin2\theta }}{g}\)

Option 3 : a parabola

**Concept:**

Equation of projectile will be of the form

y = Ax + Bx^{2}

**This is the equation of the parabola**

**Derivation:**

Let the projectile velocity be U with an angle \(\begin{array}{l} \theta \\ \end{array}\) from the horizontal

Horizontal component of velocity

\(\begin{array}{l} \mathop u\nolimits_x = u\cos \theta \\ \end{array}\)

vertical component of velocity

\(\begin{array}{l} \mathop u\nolimits_y = u\sin \theta - gt\\ \end{array}\)

\(x = u\cos \theta t\) (1)

\(y = u\sin \theta t - \frac{1}{2}g\mathop t\nolimits^2 \) (2)

substituting t from (1) in (2), we get

\(\begin{array}{l} y = x\tan \theta - \frac{{g\mathop x\nolimits^2 }}{{2\mathop u\nolimits^2 \mathop {\cos }\nolimits^2 \theta }}\\ \end{array}\)

**This is the equation of the parabola**