Question

A particle is projected at 20m/s in a direction 25 degrees above the horizontal (it has zero velocity in the vertical direction)

What maximum height does the particle reach?

What is the horizontal distance the particle has travelled when it reachrs its maximum height?

Answer 1

$$h=\frac{v^2\sin^2{25}}{2g}\\h=\frac{20^2\sin^2{25}}{2\cdot9.8}=3.65\ m\\ R=\frac{v^2\sin{(2\cdot25)}}{2\cdot g}\\R=\frac{20^2\sin{50}}{2\cdot9.8}=15.6\ m$$