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Question Part Points Submissions Used A home run is hit in such a way that the baseball just clears a wall 12 m high, located 122 m from home plate. The ball is hit at an angle of 33 degree to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.) Find the initial speed of the ball. Find the time it takes the ball to reach the wall. Find the velocity components of the ball when it reaches the wall. Find the speed of the ball when it reaches the wall. m/s

Explanation / Answer

a)

Vx = V*cos(33 degree) = 0.83867*V = 122/t

t = (122/0.83867)/V = 145.46842/V

Vy = V*sin(33 degree) = 0.544639*V

s = ut + 1/2*a*t^2

11 = 0.544639*V*145.46842/V-1/2*9.81*(145.46842/V)^2 = 0.544639*145.46842-1/2*9.81*(145.46842/V)^2

11= 0.544639*145.46842-1/2*9.81*(145.46842/V)^2

solvng we get

V = 39 m/s

b)

t = (122/0.83867)/V = 145.46842/39 = 3.73 sec

c)

Vx = V*cos(33 degree) = 0.83867*39 = 32.708 m/s ....(x- component wll be constant)

Vy = 0.544639*39-9.81*3.73 = -15.35 m/s

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