1. On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and they hit the water 1.59 s later. How high was the bridge? 2. How fast were the swimmers moving when they hit the water? 3. What would the swimmer's drop time be if the bridge were 2.5 times as high?

1) Height of bridge = 12.4 m

2) Velocity when the swimmers hit the water = 15.58 m/s

3) Time of flight if the bridge was 2.5 times as high = 2.52 s

Explanation:

Using the equations of motion,

g = 9.8 m/s², y = H = ? t = 1.59 s

Initial velocity, u = 0 m/s,

final velocity, v = ?

Total Time of fall, t = 1.59

1) y = ut + gt²/2

y = 0 + 9.8(1.59)²/2

y = 12.4 m

2) v = u + gt

v = 0 + (9.8 × 1.59)

v = 15.58 m/s

3) y = ut + gt²/2

y = 2.5 × 12.4 = 31 m

u = 0 m/s

g = 9.8 m/s²

31 = 0 + 9.8t²/2

4.9t² = 31

t² = 31/4.9

t = 2.52 s

the answer is b ) square root of 5

trig 15.3 cos