Question

A catapult fires a rock at 32 degree angle to the horizontal at a velocity of 160m/s at an oncoming vehicle. the vehicle is moving directly towards the catapult at a velocity of 35m/s. find: A) what distance should the vehicle be from the catapult at the instant the rock is released in order for the rock to hit the vehicle.
B) what is the time of flight of the rock.

Answer 1

Answer:

(a) 2347.87 m

(b) 17.3 second

Explanation:

u = 160 m/s

θ = 32°

(A) The distance of the vehicle to be hit by the rock is the horizontal range of the rock.

By use of the formula of horizontal range

$R=\frac{u^{2}Sin2\theta }{g}$

$R=\frac{160^{2}Sin2\times32 }{9.8}$

R = 2347.87 m

(B) Let T be the time o flight

Use the formula of time of flight

$T=\frac{2uSin\theta}{g}$

$T=\frac{2\times 160\times Sin32}{9.8}$

T = 17.3 second