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 124 m from home plate. The ball is hit at an angle of 37.0° 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.

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


(c) Find the velocity components of the ball when it reaches the wall.

Find the speed of the ball when it reaches the wall.

Explanation / Answer

  Vi = initial speed of ball off bat, Vx = velocity x direction, Vy = velocity y direction
Dx = distance in x direction (124m); Dy = height in y direction (10m), Di = initial height (1m)
a = acceleration due to gravity (-9.81 m/s^2), t = time

Vy = Vi(sin30) = 0.5Vi
Vx = Vi(cos30) = 0.866Vi
Dx = Vx(t)
124 = 0.866Vi(t)
143.2 = Vi(t)
Dy = -1/2at^2 + Vy(t) + Di
10 = -4.905t^2 + 0.5Vi(t) + 1
9 = -4.905t^2 + 0.5(143.2)
9 - 71.6 = -4.905t^2
62.6/4.905 = t^2
t^2 = 12.76
t = 3.57 seconds

Vi(t) = 143.2
Vi(3.57) = 143.2
Vi = 40.11 m/s

Vy = -a(t) + 0.5(Vi)
Vy = -9.81(3.57) + 0.5(40.11)
Vy = -35.02 + 20.05
Vy = -14.96 m/s

Vx = 0.866Vi
Vx = 0.866(40.11)
Vx = 34.73 m/s

Vf = sqrt(Vx^2 + Vy^2)
Vf = sqrt(1206.17 + 223.80) = 37.81 m/s
So, answers are
a) 40.11 m/s
b) 3.57 seconds
c) 34.73 m/s (x-component), -14.96 m/s (y-component)
d) 37.81 m/s

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