posted by ryan on .
A diver jumps into a 10 meter deep pool from a 5 meter platform with an initial upward velocity of 3.2 meters per second. What will the divers maximum height be?
After how many second will the diver hit the water?
If the water were drained from the pool, and I launched a ball from the platform at the same initial velocity, when would the ball hit the bottom of the pool?
You must have seen the general equation for this type of question
in your case you have
height = -4.9t^2 + 3.2t + 5
d(height)/dt = -9.8t + 3.2
= 0 for a max of height
9.8t = 3.2
t = 32/98 = 16/49 seconds
at that time,
height = -4.9(16/49)^2 + 3.2(16/49) + 5
= 5.52 m
For the second part, we would have to know how deep the pool is.
When they speak of a 10m pool, like in the Olympics, they do not refer to the depth of the pool, but rather the height of the tower for diving.
It does not mean that the pool is 10 m deep!
If the question was erroneously meant as such, then solve
-4.9t^2 + 3.2t + 15 = 0