You hit a golf ball from ground level up a hill and the ball lands 6 meters high

Question

You hit a golf ball from ground level up a hill and the ball lands 6 meters higher than where you hit it from. If the ball was initially moving at a speed of 24 m/s and left the tee at an angle of 40 degrees above the horizontal, find the following: 1. a. The time the ball was in flight before landing. b. How far the ball traveled horizontally c. The final velocity's x-component as well as the final velocity's y-component. d. The speed the ball strikes the ground with and what angle it strikes at. e. The maximum height of the ball during flight.

Explanation / Answer

(A) v0y = 24 sin40 = 15.4 m/s

ay = - 9.8 m/s^2

yf - yi = v0y t + ay t^2 / 2

6 - 0 = 15.4 t - 9.8 t^2 /2

4.9 t^2 - 15.4t + 6 = 0

t = 2.69 sec

(B) in horizontal,

x= v0x t

= 24 cos40 x 2.69

= 49.5 m

(C) vx =v0x = 18.4 m/s

vy = (24 sin40) + (-9.8 x 2.69) = - 10.9 m/s

(D) v = sqrt(vx^2 + vy^2) = 21.4 m/s

theta = tan^-1(vy / vx) = - 30.6 deg

Or 30.6 deg below the horizontal.

(E) Hmax = (v0 sin(theta))^2 / 2 g

= 12.1 m

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