The farthest a batter can hit a ball is 104.4 m. (i.e. maximum range). Suppose t

Question

The farthest a batter can hit a ball is 104.4 m. (i.e. maximum range). Suppose this batter hits another ball with the same speed as before. a. What is the launch angle, assuming the ball achieves a maximum height of 36.22 m. Assume the ball is being hit 1.2 m above the ground. b. Suppose there is a 1.2 m tall fence 98 m from the launch point. How high does the ball travel over the fence? c. What is the velocity vector and magnitude of the ball right at the moment it impacts th ground? Draw pictures...ask questions!

Explanation / Answer

1. Maximum range = v0^2 / g

104.4 = v0^2 / 9.81

v0 = 32 m/s

(A) at maximum height, vfy = 0

Applying vf^2 - v0y^2 = 2 a d

0^2 - v0y^2 = 2(-9.81) (36.22 - 1.2)

v0y = 26.2 m/s

v0y = v0 sin(theta)

theta = sin^-1 [ 26.2 / 32] = 55 deg

(B) t = 98 / (32 cos55) = 5.34 sec

y + 1.2 = (26.2 x 5.34) - (9.8 x 5.34^2 / 2)

y = 0.18 - 1.2 = -1 m

(C) vfx = 32 cos55 = 18.35

applying energy conservation,

m g h + m v0^2 /2 = m vf^2 /2

(9.81 x 1.2) + (32^2 / 2) = vf^2 /2

vf = 32.4 m/s ........Ans (magnitude)

vfy = sqrt(vf^2 - vfx^2) = 26.66 m/s

theta = - tan^-1(26.66 / 18.35) = 55.5 deg

vector = 32.4 m/s, 55.5 deg below the horizontal ....Ans

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