# Question 27767

Oct 11, 2017

See below.

#### Explanation:

The kinematic movement equations are

$\left(x , y\right) = \left({v}_{0} \cos {\theta}_{0} t , h + {v}_{0} \sin {\theta}_{0} t - \frac{1}{2} g {t}^{2}\right)$

calling $\left({x}_{1} , 0\right)$ the point of impact after ${t}_{1}$ seconds we have

$\left({x}_{1} , 0\right) = \left({v}_{0} \cos {\theta}_{0} {t}_{1} , h + {v}_{0} \sin {\theta}_{0} {t}_{1} - \frac{1}{2} g {t}_{1}^{2}\right)$

or

$\left\{\begin{matrix}{x}_{1} = {v}_{0} \cos {\theta}_{0} {t}_{1} \\ 0 = h + {v}_{0} \sin {\theta}_{0} {t}_{1} - \frac{1}{2} g {t}_{1}^{2}\end{matrix}\right.$

now solving for $h , {v}_{0}$ we obtain

{(v_0 = x_1/(costheta_0 t_1)), (h =(costheta_0 g t_1^2 - 2 sintheta_0 x_1)/(2 costheta_0)):}#

and then the speed just before the impact is computed using mechanical energy. So

$\frac{1}{2} m {v}_{0}^{2} + h m g = \frac{1}{2} m {v}_{1}^{2}$ and then

${v}_{1} = \sqrt{{v}_{0}^{2} + 2 h g}$

The item c) is left as an exercise.