Question

A girl runs at a speed of 3.9 m/s off a high dive and hit the water 1.8 s later.

a. How high was the diving board?


b. How far horizontally was she from the board when she hit the water?



c. If she had just dropped off the board, would her time to drop to the water been longer, shorter or the same?

Answer 1

Answer:

(a) the height of the diving board is 22.896 m

(b) the horizontal distance traveled by the girl is 7.02 m

(c) if she had just drop off the board, her time to drop to the water would have been longer.

Explanation:

Given;

initial speed of the girl, u = 3.9 m/s

time to hit the water, t = 1.8 s

(a) the height of the diving board is calculated as;

h = ut + ¹/₂gt²

h = (3.9 x 1.8)  +  ¹/₂ x 9.8 x 1.8²

h = 7.02 + 15.876

h =  22.896 m

(b) the horizontal distance traveled by the girl is calculated as;

X = ut

X = 3.9 x 1.8

X = 7.02 m

(c) if she just drop off the board, then the initial speed will be zero;

h = ut + ¹/₂gt²

h = 0 + ¹/₂gt²

2h = gt²

$t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2 \ \times\ 22.896 }{9.8} }\\\\t = 2.16 \ s$

Thus, if she had just dropped off the board, her time to drop to the water would have been longer.