Question

A projectile is launched with speed v0 at an angle of θ0 above the horizontal. Find an expression for the maximum height it reaches above its starting point in terms of v0, θ0, and g.

Answer 1

The projectile when was luch this has a angle repect with the horizontal, then we have two components

$v_{oy}=v_{o}.\sin \theta _{o}$

$v_{ox}=v_{o}.\cos \theta _{o}$

In the highest point, the verticaly speed is zero. Then using the equation of the moving we have the following

$v_{fy}=v_{oy}+a.t\rightarrow 0=v_{oy}-g.t\rightarrow t=\frac{v_{oy}}{g}$

Now in this equation $h=v_{oy}.t+\frac{1}{2}a.t^{2}$ replace the value of the time

$h=v_{o}.\sin \theta _{o}.(\frac{v_{o}.\sin \theta _{o}}{g})-\frac{1}{2}g.(\frac{v_{o}.\sin \theta _{o}}{g})^{2}$

$h=(\frac{v^{2}_{o}.\sin^{2} \theta _{o}}{g})-(\frac{v^{2}_{o}.\sin^{2} \theta _{o}}{2g})$

$h=\frac{v^{2}_{o}.\sin^{2} \theta _{o}}{2g}$

If you havea any question please let me know in the comments