College Physics
posted by Stephanie .
A students stands at the top of a high vertical cliff and throws calculators into the sea. He throws the calculators horizontally with a speed of 7.4 m/s and observes that each reaches the water below 3.7 s after the release. Neglect air resistance and find
(a) How far from the base of the cliff do the calculators strike he water?
(b) How high is the cliff?

Hi,
Since there is no acceleration in the horizontal direction the calculators would travel a dstance of
(horizontal velocity)X (time of flight)
=27.38
to find the height of the cliff use
D=v0t+1/2AT^2, since there is no initial velcity in the vertical direction cancel to get
D=1/2AT^2 which = 67.081
have fun!