Projectile Motion questions | Projectile Motion jee questions

Q 11 :

The angle of projection of a particle is measured from the vertical axis as $ϕ$ and the maximum height reached by the particle is $h_{m}$ . Here $h_{m}$ as function of $ϕ$ can be represented as [2025]

1

2

3

4

(3)

$h_{m} = \frac{u^{2} \sin^{2} θ}{2 g}, θ = 90 ° - ϕ$

$\Rightarrow h_{m} = \frac{u^{2} \cos^{2} ϕ}{2 g}$

From $ϕ = 0 to ϕ = 90$

$h_{m}$ decreases with $\cos^{2}$ function.

Q 12 :

A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{n u^{2}}{25 g}$ , where value of n is : (Given 'g' is the acceleration due to gravity). [2025]

6
18
12
24

1

2

3

4

(4)

Range R = $3 H_{\max}$

$\frac{u^{2} \sin 2 θ}{g} = \frac{3 u^{2} \sin^{2} θ}{2 g}$

$2 \sin θ \cos θ = \frac{3}{2} \sin^{2} θ$

$\tan θ = \frac{4}{3} \Rightarrow θ = 53 °$

$R = \frac{u^{2} (2 \times \frac{3}{5} \times \frac{4}{5})}{g} = \frac{24 u^{2}}{25 g}$

Q 13 :

Two projectiles are fired from ground with same initial speeds from same point at angles (45° + $α$ ) and (45° – $α$ ) with horizontal direction. The ratio of their times of flights is [2025]

1
$\frac{1 - \tan α}{1 + \tan α}$
$\frac{1 + \sin 2 α}{1 - \sin 2 α}$
$\frac{1 + \tan α}{1 - \tan α}$

1

2

3

4

(4)

Time of flight for $1^{st}$ projectile,

$T_{1} = \frac{2 u \sin (45 + α)}{g}$

Time of flight for $2^{nd}$ projectile,

$T_{2} = \frac{2 u \sin (45 - α)}{g}$

$\frac{T_{1}}{T_{2}} = \frac{\sin (45 + α)}{\sin (45 - α)}$

$\frac{T_{1}}{T_{2}} = \frac{\frac{1}{\sqrt{2}} \cos α + \frac{1}{\sqrt{2}} \sin α}{\frac{1}{\sqrt{2}} \cos α - \frac{1}{\sqrt{2}} \sin α}$

$\frac{T_{1}}{T_{2}} = \frac{\cos α + \sin α}{\cos α - \sin α} = \frac{1 + \tan α}{1 - \tan α}$

Q 14 :

A helicopter flying horizontally with a speed of 360 km/h at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point 'O', 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is:

(Use acceleration due to gravity g = 10 $m / s^{2}$ and neglect air resistance) [2025]

$2 \sqrt{5}$ km
4 km
7.2 km
$2 \sqrt{2}$ km

1

2

3

4

(4)

$u = 360 \times \frac{5}{18} = 100 m / s$

$x = u \times t = 100 \times 20 = 2 \times 10^{3} m = 2 km$

$t = \sqrt{\frac{2 H}{g}} \Rightarrow H = \frac{t^{2} g}{2}$

$H = \frac{400 \times 10}{2} = 2000 m = 2 km$

Displacement, $D = \sqrt{x^{2} + H^{2}} = 2 \sqrt{2} km$ .

Q 15 :

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. $T_{1}$ and $T_{2}$ are the total flying times of first and second ball, respectively, then the ratio of $T_{1}$ and $T_{2}$ is : [2025]

$2 \sqrt{2} : 1$
2 : 1
$\sqrt{2} : 1$
4:1

1

2

3

4

(1)

Given, ${(H)}_{1} = 8 \times {(H)}_{2}$

$H_{1} = \frac{u^{2} \sin^{2} θ_{1}}{2 g} and H_{2} = \frac{u^{2} \sin^{2} θ_{2}}{2 g}$

$\frac{u^{2} \sin^{2} θ_{1}}{2 g} = 8 \times \frac{u^{2} \sin^{2} θ_{2}}{2 g}$

$\sin θ_{1} = 2 \sqrt{2} \sin θ_{2} \Rightarrow \frac{\sin θ_{1}}{\sin θ_{2}} = 2 \sqrt{2}$

$T_{1} = \frac{2 u \sin θ_{1}}{g} and T_{2} = \frac{2 u \sin θ_{2}}{g}$

$\Rightarrow \frac{T_{1}}{T_{2}} = \frac{\sin θ_{1}}{\sin θ_{2}} = 2 \sqrt{2}$

Q 16 :

A particle is projected at an angle of 30° from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is $h_{0}$ and height traversed in the last second, before it reaches the maximum height, is $h_{1}$ . The ratio $h_{0}$ : $h_{1}$ is ___________ [Take, g = 10 $m / s^{2}$ ] [2025]

0%

(5)

Time to reach maximum height,

$t_{H_{\max}} = \frac{u \sin θ}{g} = \frac{60}{10} \times \frac{1}{2} = 3 s$

Height in $1^{st}$ second $S_{1} = 30 \times 1 - \frac{1}{2} \times 10 \times 1 = 25 m$

Height in last second $S_{3} = 30 + (\frac{- 10}{2}) \times (2 \times 3 - 1) = 5 m$

$\Rightarrow \frac{S_{1}}{S_{3}} = \frac{25}{5} = 5$

Q 17 :

The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance up to which he can throw the same ball is [2023]

68 m
136 m
192 m
272 m

1

2

3

4

(4)

$H_{\max} = \frac{v^{2}}{2 g} = 136 m$

$R_{\max} = \frac{v^{2}}{g} = 2 H_{\max} = 2 (136) = 272 m$

Q 18 :

The initial speed of a projectile fired from ground is $u$ . At the highest point during its motion, the speed of the projectile is $\frac{\sqrt{3}}{2} u$ . The time of flight of the projectile is [2023]

$\frac{u}{2 g}$
$\frac{u}{g}$
$\frac{2 u}{g}$
$\frac{\sqrt{3} u}{g}$

1

2

3

4

(2)

$u \cos θ = \frac{\sqrt{3} u}{2} \Rightarrow \cos θ = \frac{\sqrt{3}}{2} \Rightarrow θ = 30 °$

$Hence T = \frac{2 u \sin 30 °}{g} = \frac{u}{g}$

Q 19 :

For a body projected at an angle with the horizontal from the ground, choose the correct statement. [2023]

Gravitational potential energy is maximum at the highest point.
The horizontal component of velocity is zero at highest point.
The vertical component of momentum is maximum at the highest point.
The kinetic energy (K.E.) is zero at the highest point of projectile motion.

1

2

3

4

(1)

$At highest point V_{y} = 0$

$V_{x} = u_{x} = u \cos θ$

$U_{g} = m g h, it is maximum at H_{\max}$

Q 20 :

Two objects are projected with same velocity $' u'$ however at different angles $α$ and $β$ with the horizontal. If $α + β = 90 °$ , the ratio of horizontal range of the first object to the second object will be [2023]

1 : 2
4 : 1
2 : 1
1 : 1

1

2

3

4

(4)

$Range = \frac{u^{2} \sin 2 θ}{g}$

$Range for projection angle$ $'' α''$

$R_{1} = \frac{u^{2} \sin 2 α}{g}$

$Range for projection angle$ $'' β''$

$R_{2} = \frac{u^{2} \sin 2 β}{g}$

$α + β = 90 ° (given) \Rightarrow β = 90 ° - α$

$R_{2} = \frac{u^{2} \sin 2 (90 ° - α)}{g}$

$R_{2} = \frac{u^{2} \sin (180 ° - 2 α)}{g}$

$R_{2} = \frac{u^{2} \sin 2 α}{g}$

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{(\frac{u^{2} \sin 2 α}{g})}{(\frac{u^{2} \sin 2 α}{g})} = \frac{1}{1}$

Topic Question Set

Q 11 :

The angle of projection of a particle is measured from the vertical axis as $ϕ$ and the maximum height reached by the particle is $h_{m}$ . Here $h_{m}$ as function of $ϕ$ can be represented as [2025]

Q 12 :

A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{n u^{2}}{25 g}$ , where value of n is : (Given 'g' is the acceleration due to gravity). [2025]

Q 13 :

Two projectiles are fired from ground with same initial speeds from same point at angles (45° + $α$ ) and (45° – $α$ ) with horizontal direction. The ratio of their times of flights is [2025]

0%

Q 17 :

The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance up to which he can throw the same ball is [2023]

Q 18 :

The initial speed of a projectile fired from ground is $u$ . At the highest point during its motion, the speed of the projectile is $\frac{\sqrt{3}}{2} u$ . The time of flight of the projectile is [2023]

Q 19 :

For a body projected at an angle with the horizontal from the ground, choose the correct statement. [2023]

Q 20 :

Two objects are projected with same velocity $' u'$ however at different angles $α$ and $β$ with the horizontal. If $α + β = 90 °$ , the ratio of horizontal range of the first object to the second object will be [2023]

Want Us to Email you About Special Offers & Updates?