15.7k views
4 votes
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 2.00 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?

User Nmio
by
7.6k points

1 Answer

2 votes

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs


s=ut+0.5at^2

s=h


h=0*2+0.5*9.81*2^2\\h=19.62 meters

Part b


v=u+at\\v=0+9.81*2\\v=19.62m/s

Part c

now we have h=2*19.62=39.24


39.24=0+0.5*9.81*t^2\\t^2=8\\t=2.83 secs

User Cancan
by
8.4k points