Acrow drops .11 -kg clam onto a rocky beach from a height of 9.8 m. what is the kinetic energy of the clam when it is 5.0 m above the ground? what is it's speed at that point?

Kinetic Energy=5.17J

Speed=9.7m/sec

Explanation:

Given the mass of the clam is m=0.11kg

The height from which the rock was thrown is h=9.8m

Given the rock is at 5m above the ground.

The distance travelled by the rock is (s)=9.8-5=4.8m

The initial velocity is u=0m/sec

let the final velocity be v m/sec

We know that the acceleration of the particle is

We know that

v=9.7m/sec

Now the kinetic energy=

KE==5.17J

i think the answer would be reserve.