A student throws a baseball horizontally at 25 meters per second from a cliff 45 meters above the level ground. Approximately how far from the base of the cliff does the ball hit the ground? [neglect air resistance.]

First, calculate how long the ball is in midair. This will depend only on the vertical displacement; once the ball hits the ground, projectile motion is over. Since the ball is thrown horizontally, it originally has no vertical speed.

t = time vi = initial vertical speed = 0m/s g = gravity = - 9.8m/s^2 y = vertical displacement = - 45m

y =.5gt^2 [Basically, in this equation we see how long it takes the ball to fall 45m] - 45m =.5 (-9.8m/s^2) * t^2 t = 3.03 s

Now we know that the ball is midair for 3.03s. Since horizontal speed is constant we can simply use:

x = horizontal displacement v = horizontal speed = 25m/s t = time = 3.03s

x = v*t x = 25m/s * 3.03s = 75.76 m Thus, the ball goes about 75 or 76 m from the base of the cliff.