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 1.5s 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 was twice as high?

Answer 1

a. h = Vo*t + 4.9t^2,
h = 0 + 4.9(1.5)^2 = 11.03m.

b. Vf = Vo + gt,
Vf = 0 + 9.8*1.5 = 14.7m/s.

c. h = Vo*t + 4.9t^2 = 22.06m,
0 + 4.9t^2 = 22.06,
t^2 = 4.50,
t = 2.1s.

Answer 2

no

Answer 3

Why is 9.8 positive? Shouldn't it be negative as acceleration equals -9.8 m/s^2 and wouldn't be positive as gravity does not go up?

Answer 4

B) is -14.7. it's -9.8 m/s^2 instead of 9.8 m/s^2.