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?

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.

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

To answer these questions, we need to use the equations of motion and the concepts of free fall.

a) How high was the bridge?
To determine the height of the bridge, we can use the equation of motion for free fall:

h = 0.5 * g * t^2,

where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes for an object to fall.

Given that the swimmers take 1.5 seconds to hit the water, we can substitute t = 1.5 seconds into the equation:

h = 0.5 * (9.8 m/s^2) * (1.5 s)^2
h = 0.5 * 9.8 m/s^2 * 2.25 s^2
h = 10.87 m

Therefore, the height of the bridge is approximately 10.87 meters.

b) How fast were the swimmers moving when they hit the water?
To calculate the speed at which the swimmers hit the water, we need to use another equation of motion:

v = g * t,

where v is the final velocity, g is the acceleration due to gravity, and t is the time of free fall.

Substituting the given values:

v = (9.8 m/s^2) * (1.5 s)
v = 14.7 m/s

Therefore, the swimmers were moving at approximately 14.7 meters per second when they hit the water.

c) What would the swimmer's drop time be if the bridge was twice as high?
If the bridge height is doubled, we can use the same equation as before to find the new drop time:

h2 = 0.5 * (9.8 m/s^2) * t2^2,

where h2 is the new height and t2 is the new time of free fall.

Since the height is doubled, h2 = 2 * h = 2 * 10.87 m = 21.74 m.

Substituting the new height into the equation:

21.74 m = 0.5 * (9.8 m/s^2) * t2^2
2 * 10.87 m = 4.9 m/s^2 * t2^2
t2^2 = (2 * 10.87 m) / (4.9 m/s^2)
t2^2 ≈ 4.439 seconds^2

To find t2, we take the square root of both sides:

t2 ≈ √(4.439 seconds^2)
t2 ≈ 2.11 seconds

Therefore, if the bridge was twice as high, the swimmer's drop time would be approximately 2.11 seconds.

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?

no