A fighter plane enters inside the enemy territory, at time t=0 with velocity v0=250 m s- 1 and moves horizontally with constant acceleration a=20 m s- 2 (see figure). An enemy tank, at the border, spots the plane and fire shots at an angle θ =60° with the horizontal and with velocity u=600 m s- 1 .The altitude H , where the plane can be hit by the shot, is

Question

A fighter plane enters inside the enemy territory, at time t=0 with velocity v0=250 m s- 1 and moves horizontally with constant acceleration a=20 m s- 2 (see figure). An enemy tank, at the border, spots the plane and fire shots at an angle θ =60° with the horizontal and with velocity u=600 m s- 1 .The altitude H , where the plane can be hit by the shot, is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-qzwl78durv7z.png" /> (A) 1500√3m (B) 125 m (C) 1400 m (D) 2473 m

Tardigrade Admin · Accepted Answer

If being hit, then horizontal distance remains same.
Solution

v_{x} = 600 cos 60^{ο} = 300 m s^{- 1}

(250 m s^{- 1}) t + \frac{1}{2} (20) t^{2} = 300 t

(50 t) = \frac{1}{2} (20) t^{2}

t = 0

or

t = 5

H = ((600) (sin 60)^{ο}) t - \frac{1}{2} g t^{2}

= (600) \frac{( 3 )}{2} \times 5 - \frac{1}{2} \cdot (10) (25)

H = (3003) 5 - (125) m

H = 2473 m

$15003 m$

$125 m$

$1400 m$

$2473 m$

15003​m

125m

1400m

2473m

If being hit, then horizontal distance remains same. vx​=600 cos 60ο=300ms−1 (250 m s−1)t+21​(20)t2=300t (50 t)=21​(20)t2 t=0 or t=5 H=((600)(sin 60)ο)t−21​gt2 =(600)2(3​)​×5−21​⋅(10)(25) H=(3003​)5−(125)m H=2473m

$15003 m$

$125 m$

$1400 m$

$2473 m$