# What is the max height of the projectile motion of an object if the initial velocity was 129.98 m/s and makes angle at 24 degrees to the horizon and the total time was 10.77s?

Jun 10, 2015

$s = 142 , 6 m$.

#### Explanation:

First of all, the knowing of the "time to fly" is not useful.
The two laws of the motion are:

$s = {s}_{0} + {v}_{0} t + \frac{1}{2} a {t}^{2}$

and

$v = {v}_{0} + a t$.

But if you solve the system of the two equations, you can find a third law really useful in those cases in which you haven't the time, or you haven't to find it.

${v}^{2} = {v}_{0}^{2} + 2 a \Delta s$ in which $\Delta s$ is the space run.

It is possible to disjoint the parabolic motion in the two motion components, the vertical one (decelerated motion) and the horizontal one (uniform motion). In this exercise we only need the certical one.

The vertical component of the initial velocity is:

v_(0y)=v_0sin24°=52.87m/s.

The final velocity has to be $0$ and $a = - g$ (gravity acceleration), so:

$\Delta s = \frac{{v}^{2} - {v}_{0}^{2}}{2 a} = \frac{{0}^{2} - {52.87}^{2}}{2 \cdot \left(- 9.8\right)} = 142.6 m$.