Two balls are thrown 1 st from the tower of height 1000 m and 2 nd from 1000 3 m away from the tower on the ground with same speed of 200 m/s in same plane, then : List-I List-II (P) For θ = 5 3 ∘ (

Question

Two balls are thrown 1 st from the tower of height 1000 m and 2 nd from 1000 3 ​ m away from the tower on the ground with same speed of 200 m/s in same plane, then : List-I List-II (P) For θ = 5 3 ∘ (1) They will collide in mid air (Q) For θ = 6 0 ∘ (2) They will not collide (R) If they collide, then event will occur at height (from ground) of (3) 1154.7 m (S) Horizontal distance between point of collision and tower is (4) 833.33 m (5) 166.67 m

TestHub · Accepted Answer

The first ball travels horizontally from the tower; gravity changes both balls’ heights equally.
$1000 = 200 t sin θ \Rightarrow t = \frac{5}{s i n θ}$
For horizontal collision:
$200 t = 10003 - 200 t cos θ$
$t = \frac{5 3}{1 + c o s θ}$
The two time expressions agree for:
$θ = 6 0^{\circ}, t = \frac{10}{3} s$
Collision height and distance from tower:
$h = 1000 - \frac{1}{2} (10) (\frac{10}{3})^{2} = 833.33 m$
$x = 200 (\frac{10}{3}) = 1154.7 m$
Thus P→2, Q→1, R→4, S→3: B.

Physics - Kinematics Question with Solution | TestHub

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Doubts & Discussion

List-I	List-II
(P) For $θ = 5 3^{\circ}$	(1) They will collide in mid air
(Q) For $θ = 6 0^{\circ}$	(2) They will not collide
(R) If they collide, then event will occur at height (from ground) of	(3) 1154.7 m
(S) Horizontal distance between point of collision and tower is	(4) 833.33 m
	(5) 166.67 m