# Question 46750

Aug 24, 2014

There is no single solution. One solution is $22.944 \frac{m}{s}$ at an inclination of $59.63$ degrees.

Here are the kinematic equations that we are using:

${v}_{x} = \frac{d}{t} = \frac{40}{t}$

$10 = - \frac{1}{2} a {t}^{2} + {v}_{y} t$
$h \left(x\right) = - 4.9 {t}^{2} + {v}_{y} t - 10$

So, we have 2 equations and 3 variables: ${v}_{x}$, ${v}_{y}$, and $t$. This is why there is no unique solution.

Since it hasn't been specified, we will assume the ball is being thrown from the ground, $0$ $m$.

If we want to be realistic, there should be some arc to the ball so that it won't hit the rim on the way down. So, let's make the ball peak at $20$ $m$ (you can change this if you like). This will give us ${v}_{y}$ and only 2 variables remaining.

$20 = \frac{1}{2} a {t}_{p e a k}^{2}$
$20 = 4.9 {t}_{p e a k}^{2}$
${t}_{p e a k}^{2} = \frac{20}{4.9}$
${t}_{p e a k} = 2.02$ $s$
${v}_{y} = 9.8 \cdot 2.02 = 19.796 \frac{m}{s}$

Now let's solve the vertical component:

$h \left(x\right) = - 4.9 {t}^{2} + {v}_{y} t - 10$
$= - 4.9 {t}^{2} + 19.796 t - 10$

We can put this into our calculator and solve the roots or you can solve with the quadratic formula:

$t = .5912$ $s$
$t = 3.4482$ $s$

We reject the first value, because the ball is going up and hasn't reached the net horizontally. With our time value, we can now calculate the horizontal velocity:

${v}_{x} = \frac{40}{3.4482} = 11.6 \frac{m}{s}$

Now we have to answer the question. The velocity can be solved with Pythagorean and the angle can be solved with trigonometry.

v=sqrt(v_x^2+v_y^2) #=sqrt(11.6^2+19.796^2)
$= 22.944 \frac{m}{s}$

$\theta = {\tan}^{- 1} \frac{{v}_{y}}{{v}_{x}}$
$= {\tan}^{- 1} \frac{19.796}{11.6}$
$= 59.63$ degrees