A golf ball is dropped from rest from a height of 9.40 m. It hits the pavement,

Question

A golf ball is dropped from rest from a height of 9.40 m. It hits the pavement, then bounces back up, rising just 5.90 m before falling back down again. A boy then catches the ball when it is 1.10 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch

I got 2.91s, but that was wrong. I assumed drop A had a Vo=o m/s, a=-9.80 m/s^2, y=-9.40 m and used V^2=V0^2+2ay to find the final velocity and y=Vot+1/2at^2 to find the time. I then used the final velocity of the first drop as the initial velocity of bounce B and found time. Then I used Vo=o m/s as the initial velocity for the drop B and found time. What am I doing wrong, and how do i go about this? Please explain!

Explanation / Answer

Let's start by finding the velocity just like you did.

Time taken can be calculated as,

Now we can use the above equation again to find the time. We will get a quadratic in t, with 2 solutions, for when the ball is passing height hc on the way up and on the way down. We are looking for the larger number of the 2.

The 2 solutions are,

2.69 seconds is the one we are looking for.

Therefore the total time would be,

Hope this helps. Good luck!

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