Question

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

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.