Solve Perplexing Physics Problems with These Helpful Tips and Examples"

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To solve the first physics problem, the second rock must be dropped before the first rock reaches its peak height, as it will have a downward speed of 15 m/s when passing the initial height again. The total time for both rocks to hit the ground must be calculated to determine the correct timing for dropping the second rock. For the second problem, average acceleration can be found using the formula for acceleration, which is the change in velocity divided by the time of contact with the wall. The key is to remember that the ball's velocity changes from 25 m/s to -22 m/s (considering direction) during the 3.5 ms contact time. Understanding these concepts will help in solving both problems effectively.
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Hi.
I'm having problems with some Physics problems; among them:

9. At the top of a cliff a 100 m high, a rock is thrown upward with velocity 15 m/s. How much later should a second rock be dropped from rest so both stones arrive simultaneously at the bottom of the cliff?

a. 5.05 s

b. 3.76 s

c. 2.67 s

d. 1.78 s

2. A 50-gram superball traveling at 25 m/s is bounced off a brick wall and rebounds at 22 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.5 ms, what is the average acceleration of the ball during this time interval?

a. 13,428 m/s2

b. 6,715 m/s2

c. 857 m/s2

d. 20 m/s2

For 9), I thought I'd use v = u + at and find 2t after solving:
0 = 15 - 9.8t for t, and that would give me the answer... but I'm wrong.
Anyone know how I can go about this one?
And for 2), I really don't know what to do...

Thanks for any help.
 
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For the first one, your answer was wrong because the first rock will have a downward speed of 15 m/s when it passes you again, and the other rock is dropped from rest. You need to drop the second rock before this point, and the only way to figure out when is to find the total time it will take each rock to hit the ground.

For the second, acceleration is the change in velocity over time. This problem is easier then it seems.
 
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