7: A pebble is dropped from rest from the top of a tallcliff and falls 4.9 m after 1.0 s has elapsed. How much fartherdoes it drop in the next 2.0 s?(a) 9.8 m (b) 19.6 m (c) 39.0 m (d) 44.0 m (e) NoneQ18: A pebble is released from rest at a certain height andfalls freely, reaching an impact speed of 4 m/s at the floor.Next, the pebble is thrown down with an initial speed of 3m/s from the same height. What is its speed at the floor?(a) 4 m/s (b) 5 m/s (c) 6 m/s (d) 7 m/s (e) 8 m/sQ19: A ball is thrown straight up in the air. For which situ-ation are both the instantaneous velocity and the accelerationzero?(a) On the way up(b) At the top of its flight path(c) On the way down(d) Halfway up and halfway down(e) None of the aboveQ20: A 50.0 g ball traveling at 25.0 m/s bounces off a brickwall and rebounds at 22.0 m/s. A high speed camera records8

Sure, I can help with that. Here are the answers to your questions:

Q7: The distance an object falls under gravity is given by the equation d = 0.5gt^2, where g is the acceleration due to gravity (9.8 m/s^2) and t is the time. After 1 second, the pebble has fallen 4.9 m. In the next 2 seconds, it will fall an additional 0.59.82^2 = 19.6 m. So the answer is (b) 19.6 m.

Q18: The final speed of an object falling under gravity is given by the equation v = u + gt, where u is the initial speed, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. If the pebble reaches an impact speed of 4 m/s when dropped from rest, then 4 = 0 + 9.8t. Solving for t gives t = 4/9.8 = 0.41 s. If the pebble is thrown down with an initial speed of 3 m/s, then its final speed will be v = 3 + 9.80.41 = 7 m/s. So the answer is (d) 7 m/s.

Q19: The instantaneous velocity of a ball thrown straight up in the air is zero at the top of its flight path, but the acceleration is not zero - it is equal to -g (the acceleration due to gravity). So the answer is (b) At the top of its flight path.

Q20: This question seems to be incomplete. Could you please provide the rest of the question?