A man standing 28 m from the base of a vertical cliff throws a ball with a speed

Question

A man standing 28 m from the base of a vertical cliff throws a ball with a speed of 33 m/s aimed directly at a point 6 m above the base of the cliff.

a) How long does it take the ball to reach the cliff?

b) Neglecting air resistance and the height of the man, calculate the height above the base of the cliff at which the ball hits.

c) How fast is the ball moving when it reaches the cliff?

d)At what time does the ball reach its largest vertical height?

e)At what horizontal distance from the man does the ball reach its largest vertical height?

Explanation / Answer

a)28=33*28/28.635 t t=0.867 sec b)y=33*6/28.635*0.867-0.5*9.8*0.867*0.867 =2.311 m above the base of cliff c)Vx=33*28/28.635=32.628 m/sec Vy=33*6/28.635-9.8*0.867 =-1.582 m/sec speed=32.66 m/sec d)t=33*6/28.635*(1/9.8) =0.705 sec e)x=33*28/28.635*0.705 =22.749 m

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