Question

Mr. Martin springs off a 5 meter high diving board with an upward velocity of 20 m/s.
A. Find Mr. Martin's height, velocity and acceleration after 0.5 seconds.

B. Determine when Mr. Martin will hit the water.

C. What is Mr. Martin's maximum height off the board?

D. What is Mr. Martin's maximum velocity when he hits the water?

E. What is Mr. Martin's velocity when he reaches half of his maximum height?

Answer 1

v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2

A.
after .5 seconds:
v = 20 - 9.81(.5)
h = 5 + 20 (.5) - 4.9 (.25)

B.
0 = 5 + 20 t - 4.9 t^2
solve quadratic equation for t when h is zero

C.
at top v = 0
0 = 20 - 4.9 t at top
use t at top for height
h = 5 + 20 t - 4.9 t^2

D.
v = 20 - 8.81 t
(it better be negative using the t from part B )

E.
use (1/2) the part C answer (I guess, it does not say off the water or off the board) for h in
h = 5 + 20 t - 4.9 t^2
get t then use it in
v = 20 - 9.81 t

of course there may be 2 answers for t, on the way up and on the way down.

Answer 2

By the way 20 meters per second is quite a spring. Are you sure it is not 2 ?
say you jump with a speed up of 20 m/s from the ground
at top. I will call g = 10 instead of 9.81

0 = 20 - 10 t
t = 2 seconds upward

h = 5 t^2 = 5(4) = 20 meters
that is like 20 yards
60 feet !!!
good grief.