Question

Suppose you throw the 0.12-kg ball at this minimum speed. Your friend misses the catch. What is the ball's kinetic energy just before it hits the ground (2 s.f.)?

Answer 1

You friend is 11 m above you

minimum speed is u. tha ball reaches your friend with zero speed.

using $v^2-u^2=2as$

$0-u^2=-2\times9.81\times11$

$u=14.69\,m/s$

kinetic energy of ball at bottom = Initial kinetic energy of ball + Initial potential energy

$\frac{1}{2}mv^2=\frac{1}{2}mu^2+mgh$

$\frac{1}{2}mv^2=0.12\times9.81\times11+0.12\times9.81\times12=27.1\,J$