Suppose a skydiver jumps out of a plane at 15,000 meters above the ground. it takes him 2.0 seconds to pull the cord to deploy the parachute and another 2.0 seconds for the parachute to be fully deployed. additionally, in order to land safely on the ground, his parachute must be fully deployed 10.0 seconds before he hits the ground. with this information and assuming there is no additional air resistance or changes in atmospheric density, what is the maximum time into his fall that he can wait to pull the cord in order to make it safely to the ground?

The time he can wait to pull the cord is 41.3 s

Explanation:

The equation for the height of the skydiver at a time "t" is as follows:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

First, let´s calculate how much time will it take for the skydiver to hit the ground if he doesn´t activate the parachute.

When he reaches the ground, the height will be 0 (placing the origin of the frame of reference on the ground). Then:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 15000 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

0 m = 15000 m - 4.9 m/s² · t²

-15000 m / -4.9 m/s² = t²

t = 55.3 s

Then, if it takes 4.0 s for the parachute to be fully deployed and the parachute has to be fully deployed 10.0 s before reaching the ground, the skydiver has to pull the cord 14.0 s before reaching the ground. Then, the time he can wait before pulling the cord is (55.3 s - 14.0 s) 41.3 s.

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures