A home run is hit in such a way that the baseball just clears a wall 10.0 m high

Question

A home run is hit in such a way that the baseball just clears a wall 10.0 m high, located 122 m from home plate. The ball is hit at an angle of 33.0° 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. x-component m/s y-component m/s Find the speed of the ball when it reaches the wall. m/s

Explanation / Answer

Ball is hit 33° to the horizontal so the relation between the x and y components of the initial velocity is:
tan 33° = vy / vx
vy = vx*tan 33°

vx is constant, however vy changes with time.

x = 122m
y = 9m (10m - 1m)

x = vx*t
y = vy*t +1/2*g*t2

now first we have

x = vx*t
122m = vx*t
vx = 122m / t

Substitute for vy then vx:
9m = vx * tan 33° * t + 1/2 * -9.8m/s2 * t2
9m = 122m / t * tan 33° * t + 1/2 * -9.8m/s2 * t2
9m = 79.22 – 4.9* t2
t = 3.78 s

a)
cos 33° = vx / velocity
vx = 122m / 3.78 s

vx = 32.27 m/s

v = vx / cos 33°

v = 32.27 m/s / cos 33°

v = 38.47 m/s

b)

t = 3.78 s

c)

vx = 32.27 m/s

Initial vy = 38.47 m/s * sin 33°

Initial vy = 20.95 m/s

Final vy = 20.95 m/s + (-9.8m/s2) * 3.78 s

Final vy = -16.1 m/s

v = (vx2 + vy2)

v = ((32.27)2 + (-16.1 m/s)2)

v = 36.06 m/s

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