a person jumps from a window 15m above a safety net. The survivor stretches the net 1.0 m before coming to rest. What was the average deceleration experienced by the person when they were slowed to rest by the net? I do know the answer is -150m/s2,but I don't know how the answer was found.
Find the velocity when the person hits the net, a free fall of 15 m.
Vfinal= sqrt (2gh) where g is the acceleration due to gravity, and h is the height.
Now the physics of the net.
The average velocity will be the initial velocity (above) divided by two. So the time it takes in the net is (time=1m/avgvelocity).
Finally,
avg deacceleration=change in velocity/time
= vabove/time=sqrt(2gh)/1m/avgvel)
= sqrt(2gh)vavg /1=sqrt(2gh)sqrt(2gh)/2
= gh
Neat question. Check my thinking.
Your thinking and approach to solving the problem is correct!
To find the velocity when the person hits the net, we can use the equation Vfinal = sqrt(2gh), where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height (15 m). Plugging in the values, we get Vfinal = sqrt(2 * 9.8 m/s^2 * 15 m) ≈ 17.146 m/s.
Next, we need to find the time it takes for the person to go through the net, which stretches 1.0 m. Since the average velocity is equal to the initial velocity divided by two, we can calculate the time using the formula time = distance / average velocity. In this case, the distance is 1.0 m and the average velocity is 17.146 m/s, so the time it takes is approximately 0.0582 seconds.
Finally, we can find the average deceleration experienced by the person when they were slowed to rest by the net. The average deceleration is equal to the change in velocity divided by time. Since the initial velocity is 17.146 m/s and the final velocity is 0 m/s (because the person is brought to rest), the change in velocity is -17.146 m/s. Dividing this by the time of 0.0582 seconds, we get an average deceleration of approximately -150 m/s^2.