Question

Two divers jump from a 3-m platform. One jumps upward at 1.50m/s , and the second steps off the platform as the first passes it on the way down.

a) What is the first diver's speed as he hits the water?

b) What is the second diver's speed as he hits the water?

c) Which hits the water first?

Answer 1

Working formula is

Vf^2 - Vo^2 = 2gs

where

Vf = speed at which diver will hit the water
V = initial speed of the diver
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
s = distance of platform from the water

For the first diver,

Vf\^2 - 1.5^2 = 2(9.8)(3)

Vf^2 = 2.25 + 2(9.8)(3)

Vf^2 = 61.05

Vf = 7.81 m/sec.

For the second diver,

Vf^2 + 0 = 2(9.8)(3)

Vf^2 = 7.67 m/sec

Diver speeds:

First diver -- 7.81 m/sec.
Second diver -- 7.67 m/sec.

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Working formula

Vf - Vo = gT

where

T = time for diver to reach the water

and all the other terms have been previously defined.

Substituting values,

For the first diver,

3 = 1.5T + (1/2)(9.8)T^2

Rearranging,

4.9T^2 + 1.5T - 3 = 0 and using the quadratic formula,

T = 0.644 sec.

and for the second diver,

3 = 0 + (1/2)(9.8)T^2

T^2 = 6/9.8

T = 0.782

Divers' times to hit the water from the platform:

First diver --- 0.644 sec.
Second diver -- 0.782

Since T for the first diver is less than the second diver, the first diver will reach the water first