A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle

Question

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60 degree above horizontal. It lands on the edge of the cliff 4.0 s later. How high is the cliff? Express your answer to two significant figures and include the appropriate units. What was the maximum height of the ball? Express your answer to two significant figures and include the appropriate units. What is the ball's impact speed? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

speed of the ball in vertical direction Vv= 33sin(60 degree) = 28.57 m/s

and in horizontal direction Vh =33cos(60 degree)= 16.5 m/s

part (B) maximum height of the ball Hmax = (u^2)/2g

where u is initial speed in vertical direction and g = 9.8 m/s^2 acceleration due to gravity

=> Hmax = ((28.57)^2)/(2*9.8) =41.66 m

time taken to reach max height is t = u/g = 28.57/9.8 = 2.91 sec

=> from max. height point after t=(4.00-2.91)= 1.09 sec ball strik cliff

PART (A) vertical distance travelled by ball from max height point to strik point

d = ut +(at^2)/2 = 0 + [(9.8)*(1.09)^2]/2 = 5.82 m

=> HEIGHT OF CLIFF = Hmax - d = 41.66 - 5.82 = 35.84 m

PART (C) Vh = 16.5 m/s will remain same

calculating from max height point where u = 0

Vv = u +at = 0 + (9.8)*(1.09) = 10.68 m/s

SO impact speed V = [Vh^2 + Vv^2]^0.5 = [272.25 + 114.06]^0.5 = 19.65 m/s

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