A home run is hit in such a way that the baseball just clears a wall 10 m high,

Question

A home run is hit in such a way that the baseball just clears a wall 10 m high, located 136 m from home plate. The ball is hit at an angle of 37° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.)

(a) Find the initial speed of the ball. m/s


(b) Find the time it takes the ball to reach the wall.   s

(c) 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

theta = 37 degrees

t = 136 / u cos(theta)

t = 136/ u cos(37 degrees)

10 = 1 + u sin(37 degrees) t - 0.5 * g * t^2

9 = 136* tan(37 degrees) - 4.905 * (136/u)^2 * sec^2(37 degrees)

Solving above equation, we get u = 39 m/s

t = 136/( u cos(37 degrees)) = 136 / ( 39 * cos(37 degrees)) = 4.366 s

(a) Initial speed =  39 m/s

(b) Time taken = 4.366 s

(c) X component = u cos(theta) = 39 * cos(37 degrees) = 31.15 m/s

Y component = u sin(theta) - g t = 39 * sin(37 degrees) - 9.81 * 4.366 = - 19.356 m/s

(d) speed = sqrt(31.15^2 + 19.356^2) = 36.67 m/s

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