Question

A sniper fires a bullet at 120 m/s at 30° above the horizontal from the roof top of a 35 m high parking garage. If the bullet strikes the level ground beside the parking garage: a. How long was the bullet in the air? b. How far from the base of the parking garage did the bullet land? c. At what angle did the bullet land?

Answer 1

Let, the bullet was there in the air for t seconds vertical distance covered by the bullet h = -35 m h = (V_0 sin theta) t - 1/2 gt^2 [Because acceleration in the vertical direction = -g] -35 = (120 sin 30 degree) t - 1/2 times 9.81 times t^2 (4.905 t^2 - 60 t - 35) = 0 Solution of the above equation yields t = -0.558 s, 12.79 sec negative value of t is not acceptable Therefore t = 12.79 sec Therefore The bullet was there in the air for 12.79 sec \r\n distance of the point where the bullet has landed from the parking garage, R = (V_0 cos theta)^t [Because acceleration in the horizontal direction = 0] = [120 cos (30 degree) times 12.79] = 1329.18 m y-component of velocity at the landing points V_y = V_0 sin theta - gt = (120 sin (30 degree) - 9.81 times 12.79) = -65.469 m/s x-component of velocity at the landing points

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