Question

A diver runs and dives off the end of a spring board that is 0.9 meters above the surface of the water. As she takes off from the diving board, her velocity is 5.2 m/s directed 36° above horizontal calculate:

a) the time she is in the air

s
b) the distance she lands from the board (measured horizontally from the end of the board) m
c) the maximum height she attains above the surface of the water

m

Answer 1

the velocity v comprises x- and y- components:

vx = 5.2 cos36° = 4.2
vy = 5.2 sin36° = 3.1

y = 0.9 + 3.1t - 4.9t^2
y=0 when t=0.85
so, she is in the air 0.85 seconds

x = 4.2t
4.2(0.85) = 3.57m
so, she lands 3.57m from the end of the board.

y is max when t = -3.1/-9.8 = 0.32
y(.32) = 1.39
max height is thus 1.39m above the water