Dive right in!
(Assume a value for the Earth's gravitational acceleration of exactly vertically downwards)
Al and Bill have had a deep swimming pool constructed, with a diving platform and a diving board both at precisely 16 feet above the water. Al climbs onto the diving platform and dives into the pool. He begins with zero vertical motion, and thus takes exactly one second to reach the pool's surface.
Bill climbs onto the diving board, prepared to launch himself upwards a bit, thus arcing towards the pool. He does so and takes exactly 2 seconds to reach the pool's surface.
What was Bill's maximum height above the water, in feet?
The answer is 25.
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1 solution
First we need to find Bill's initial velocity.
Using the standard equation, displacement (-16 feet) is equal to v t + ( 1 / 2 ) a t 2
t = 2 s e c and a = − 3 2 f t / s e c 2 , so v = 2 4 f t / s e c
So Bill reaches his peak height at (24/32) = 3/4 of a sec after takeoff.
Substituting t = 3/4, we get: d = ( 2 4 ) ( 3 / 4 ) + ( ( 1 / 2 ) ( − 3 2 ) ( ( 3 / 4 ) 2 ) ) = 9 ft
Since the initial height was 16 ft. 9 + 16 = max height of 25 ft.