A boy kicks a ball horizontally near the edge off a boardwalk, with an initial s

Question

A boy kicks a ball horizontally near the edge off a boardwalk, with an initial speed of 9.0 m/s. A blowing wind gives the ball a constant horizontal acceleration of 12/m/s^2. The ball falls into the water directly under the boy. Imagine the effect of air resistance on a the vertical motion of the ball.

a) Determine the height of the boardwalk above the water.

b) If the blowing wind reverse direction while maintaining the same strength, where does the ball fall when it is kicked with the same initial speed?

Explanation / Answer

(a) the displacement of ball in horizontal is 0

0 = 9t - 1/2*12t^2

t = 3/2 = 1.5

this is time taken by the ball to fall in vertical

h = 1/2*g*t^2

= 0.5*9.8*(1.5)^2 = 11.025m

(b) if air reverses direction

then time taken by ball to fall will be the same = 1.5 s

But distance in horizontal axis will change

s = 9t + 6t^2

s = 27m

So the ball will fall 27m away from the boy.

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