A home run is hit in such a way that the baseball just clears a wall 14.0 m high
ID: 1657600 • Letter: A
Question
Explanation / Answer
The ball is hit 33 degree to the horizontal so the relation between the x and y components of the initial velocity is:
tan (33 deg) = vy / vx
vy = vx * tan (33 deg)
vx is constant, but vy changes with time.
delta(x) = 118 m
delta(y) = 13m (14m - 1m)
Therefore:
delta(x) = vx*t
delta(y) = vy*t +1/2*g*t^2
Plug in known values:
118 m = vx*t
vx = 118m / t
Substitute for vy then vx:
13 m = vx * tan(33)* t + 1/2 * -9.81m/s^2* t^2
13 m = 118 m / t * tan(33 deg)* t + 1/2 * -9.81m/s^2* t^2
The t cancel out:
13m - 118m*tan (33 deg) = 1/2 * -9.81m/s^2 * t^2
t = sqrt((13m - 118m*tan (33 deg)) / (1/2 * -9.81m/s^2))
t = 3.6017 s
The ball travels 3.6017 s before going over the fence
(a) We can use the constant x component of the velocity to calculate the initial velocity.
The relation between the velocity and its x component is:
cos (33 deg) = vx / velocity
vx = dx / t = 118m / 3.6017 s = 32.76 m/s
velocity = vx / cos (33 deg) = 32.76 m/s / cos (33 deg) = 39.06 m/s
(b) Already calculated: t = 3.6017 s
(c) vx is constant at 32.76 m/s
vy can be calculated using the equation:
v = v0 + a*t
Initial vy = 39.06 m/s * sin (33 deg) = 21.27 m/s
Final vy = 21.27 m/s + (-9.81m/s^2) * 3.6017 s
Final vy = -14.06 m/s
The speed when the ball reaches the wall is (using Pythagorean Theorem):
v = sqrt(vx^2 + vy^2)
v = sqrt((32.76 m/s)^2 + (-14.06 m/s)^2)
v = 35.65 m/s
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