Question

1. A ball of 550 grams is kicked at an angle of 25° with the ground with an initial velocity Vi.
a) What is the initial velocity V0 of the ball if its kinetic energy is 25 Joules when its height is maximum?
b) What is the maximum height reached by the ball?

Answer 1

a) At the highest point, the ball has only horizontal component of velocity, which is

$$v_x=v_0\text{cos}\theta.$$

The kinetic energy at this point is

$$KE=\frac12 mv_x^2=\frac12mv_0^2\text{cos}^2\theta,\\\space\\ v_0=\sqrt{\frac{2KE}{m\text{ cos}^2\theta}}=10.5\text{ m/s}.$$

b) The maximum height of a ball thrown at an angle above the horizontal is

$$h=\frac{v_y^2}{2g}=\frac{v_0^2\text{sin}^2\theta}{2g}=1\text{ m}.$$