From a cliff 37.6 m high. At the level of the sea, a rock sticks out a horizontal distance of 12.12 m. The acceleration of gravity is 9.8 m/s 2 . With what minimum horizontal velocity must the cliff divers leave the top of the cliff if
they are to miss the rock? Answer in units of
m/s.

4.3 m/sec

Explanation:

Here height of cliff = y = 37.6 m

Gravitational acceleration = g = 9.8 m/sec2

vi = 0 m/s

Let's find the time which the diver will take if jumps from there!

Using formula

y = vit+1/2gt2

==> 37.6= 0 + 0.5 ×9.8×

==>=

==> t = 2.8 sec

In this time the diver has to cover a horizontal distance of 12.12 m

If x = 12.12 m is the horizontal distance to be covered then using

x= Vx × t

==> Vx = x/t

==> Vx= 12.12/2.8 = 4.3 m/s

According to Newton's third law, the force exerted by the bat hitting the ball will be equal in magnitude but opposite in direction of the force the ball exerts on the bat. Generally, your arms are stiff when you hit the ball forward, so you will not feel the bat "recoiling"

Explanation: