Physics help please!?

Question

please answer one or more of these questions:

1)A climber near the summit of a vertical cliff accidentally knocks loose a large rock. It shatters at the bottom 8s later. What was the speed of impact? How far did the rock fall?

2)A ball is thrown with enough speed straight up so that it is in the air for several seconds. (a) What is its velocity 1s before it reaches its highest point? (b) What is its velocity 1s after it reaches its highest point? (c) What is the change in velocity during the 2s interval? (d) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?
(this one I just need help with d)

3) Suprisingly, very few athletes can jump more than 2 feet straight up. Use d=1/2*gtt and solve for the time one spends moving upwards in a 2-foot vertical jump.

PLEASE HELP!!!

Answer 1

1. average velocity = d/8 m/s
gravity = 9.8 m/s^2

Speed of impact = 9.8*8
=78.4 m/s

Average velocity = 78.4/2=d/8

d=78.4*4
=313.6m

2. Part d the acceleration is always constant at zero velocity sice the earth's gravity is the only force acting on it. 9.8 m/s^2

3. using 32ft/s^2 for gravity

2=16t^2

t=sqrt(2)/4

Answer 2

N.1
Use the formula v=u+at where v is the final speed and u the initial speed.
You know the initial speed (0m/s), the acceleration (9.8m/s) and the time (8s) therefore the final speed is 0 + 9.8x8
N.2 part d
Acceleration is constant, so at all points it is 9.8m/s2, the ball has 0m/s velocity at the top point in its path.

N.3
Rearrange your formula so t is the subject. (2d/g)=tt. This will give you the time for the person to rise to the top point or fall to the ground again, therefore you must x2 to get the total time in the air.
Hope that helps!