# Question 26de5

##### 1 Answer
Mar 30, 2015

The answer are:

${v}_{0 x} = 17.2 \frac{m}{s}$,

${v}_{0 y} = 24.5 \frac{m}{s}$,

final velocity $v = 30 \frac{m}{s}$,

final angle theta=125°.

The motion of a projectile is a composition of two indipendent motions. The first, horizontal, is an uniform motion ($\vec{v}$ is costant), and the second, vertical, is an accelerated motion (with $a = - g$, $g$ is the gravity acceleration).

So we can consider, for now, the vertical motion. The low is:

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

that becomes, in this case,

$v = {v}_{0} - g t$.

The projectile goes up until his velocity is $0$, and the time of fly is $2.5 s$, so:

${v}_{0} = v + g t = 0 + 9.8 \frac{m}{s} ^ 2 \cdot 2.5 s = 24.5 \frac{m}{s}$,

that is the vertical component we are searching.

${v}_{0 y} = 24.5 \frac{m}{s}$.

Given $\theta$ is the angle of ${\vec{v}}_{0}$ with respect to the horizontal, we know that:

tantheta=v_(0y)/v_(0x)rArrv_(0x)=v_(0y)/tanthetarArrv_(0x)=(24.5m/s)/tan(55°)~=17.2m/s.

So:

${v}_{0} = \sqrt{{24.5}^{2} + {17.2}^{2}} \cong 30 \frac{m}{s}$.

When the projectile lands, his velocity is $v = 85 \frac{m}{s}$, and the angle with respet of the horizontal is theta=180°-55°=125°#, because the motion is specular.