Question

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 I estimated that they hit the water 1.50 s later.

A.) How high was the bridge?

B.) How fast were the swimmers moving when they hit the water?

C.)What would the swimmer's drop time be if the bridge were twice as high?

Answer 1

Let us consider the downward direction as positive.

Gravitational acceleration = g = 9.81 m/s²

Initial velocity of the swimmer's = V₀ = 0 m/s

Time taken to hit the water = T = 1.5 sec

Height of the bridge = H

H = VT + gT²/2

H = (0)(1.5) + (9.81)(1.5)²/2

H = 11.03 m

Speed of the swimmer's when they hit the water = V₁

V₁ = V₀ + gT

V₁ = 0 + (9.81)(1.5)

V₁ = 14.71 m/s

New height of the bridge = H₁ = 2H = 22.06 m

Time taken by the swimmer's to hit the water after stepping off the new bridge = T₁

H₁ = V₀T₁ + gT₁²/2

22.06 = (0)T₁ + (9.81)T₁²/2

T₁ = 2.12 sec

A) Height of the bridge = 11.03 m

B) Speed of the swimmer's when they hit the water = 14.71 m/s

C) Swimmer's drop time if the bridge were twice as high = 2.12 sec