# Question ae3f1

Aug 5, 2017

${v}_{0} = 28.0$ $\text{m/s}$

#### Explanation:

We're asked to find the speed at which the ball was hit, given its horizontal range and hit angle.

To do this, we can use the equations

ul(Deltay = v_0sinalpha_0t - 1/2g t^2

ul(Deltax = v_0cosalpha_0t

where

• $\Delta y$ is the change in height ($0$ since the ball landed at the same height it was hit)

• $\Delta x$ is the change in horizontal position, given as $103$ $\text{m}$

• ${v}_{0}$ is the initial speed (what we're trying to find

• ${\alpha}_{0}$ is the hit angle, given as ${50}^{\text{o}}$

Plugging in known values to both equations:

$0 = {v}_{0} \sin \left({50}^{\text{o")t - 1/2(9.81color(white)(l)"m/s}} ^ 2\right) {t}^{2}$

 103color(white)(l)"m" = v_0cos(50^"o")t

We can realize that the time $t$ when the ball is at $y$-position $0$ and $x$-position $103$ $\text{m}$ are equal, so we can solve the second equation for $t$ and plugbthat in for $t$ in the first equation (and eliminate units to make it simpler):

$t = \frac{103}{{v}_{0} \left(0.643\right)} = \frac{160.2}{{v}_{0}}$

$0 = {v}_{0} \left(\frac{160.2}{{v}_{0}}\right) - \frac{1}{2} \left(9.81\right) {\left(\frac{160.2}{{v}_{0}}\right)}^{2}$

Now we solve for ${v}_{0}$:

$4.905 {\left(\frac{160.2}{{v}_{0}}\right)}^{2} = 160.2$

$\frac{125944}{{\left({v}_{0}\right)}^{2}} = 160.2$

v_0 = sqrt((125944)/(160.2)) = color(red)(ulbar(|stackrel(" ")(" "28.0color(white)(l)"m/s"" ")|)#