Posted by **Stephanie** on Monday, October 10, 2011 at 9:05pm.

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?

- College Physics -
**Shoa**, Thursday, October 13, 2011 at 8:10pm
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!

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