Question

Question #46750

Answer 1

There is no single solution. One solution is `22.944m/s` at an inclination of `59.63` degrees.

Here are the kinematic equations that we are using:

`v_x=d/t=40/t`

`10=-1/2at^2+v_yt`
`h(x)=-4.9t^2+v_yt-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=1/2at_(peak)^2`
`20=4.9t_(peak)^2`
`t_(peak)^2=20/4.9`
`t_(peak)=2.02` `s`
`v_y=9.8*2.02=19.796m/s`

Now let's solve the vertical component:

`h(x)=-4.9t^2+v_yt-10`
`=-4.9t^2+19.796t-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=40/3.4482=11.6m/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)
`=22.944m/s`

`theta=tan^(-1)(v_y)/(v_x)`
`=tan^(-1)(19.796)/(11.6)`
`=59.63` degrees