A boy kicks a ball horizontally near the edge of a boardwalk, with an initial sp

Question

A boy kicks a ball horizontally near the edge of a boardwalk, with an initial speed of 9.0 m/s. A blowing wind gives the ball a constant horizontal acceleration of 12 m/s2. The ball falls into the water directly under the boy. Ignore the effect of air resistance on the vertical motion of the ball.

Determine the height of the boardwalk above water.

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

Please be as detailed as you can...Thanks

Explanation / Answer

Along X-direction :

Vi = initial velocity = 9 m/s

ax = acceleration = -12 m/s2

dx = displacement in X-direction =0

t = time taken

Using the formula

dx = Vi t + (0.5) ax t2

0 = 9 t + (-6) t2

t = 1.5 sec

Along vertical direction :'

Vi = initial velocity = 0 m/s

ay = acceleration = 9.8 m/s2

dy = displacement in Y-direction

t = time taken

Using the formula

dy = Vi t + (0.5) ay t2

dy = (0) (1.5) + (0.5) (9.8) (1.5)2

dy = 11.025 m    height of boardwalk

distance travelled in horizontal direction :

Using the formula

dx = Vi t + (0.5) ax t2

dx = 9 (1.5) + (0.5) (6) (1.5)2

dx = 20.25 m from boy

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