Question

Projectile motion question?

A basketball leaves a player's hands at a height of 2.10 m above the floor. The basket is 2.60 m above the floor. The player likes to shoot the ball at a `38.0^@` angle. If the shot is made from a horizontal distance of 11.00 m and must be accurate to `+-0.22 m` (horizontally), what is the range of initial speeds allowed to make the basket?

Thanks for your help

Answer 1

`v_0 in [10.8color(white)(l)"m/s", 11.0color(white)(l)"m/s"]`

Explanation:

WARNING: Long answer!

We're asked to find the range of initial speeds that a player can make a shot, with some given measurements.

Let's take a look at our known quantities:

• initial height `y_0` is `2.10` `"m"`

• final height `y` is `2.60` `"m"`

• angle of projection `alpha_0` is `38.0^"o"`

• the minimum horizontal range `Deltax_"min"` is `11.00` `"m"` `- 0.22` `"m"` `= 10.78` `"m"`

• the maximum horizontal range `Deltax_"max"` is `11.00` `"m"` `+ 0.22` `"m"` `= 11.22` `"m"`

What we can do to solve this problem is equate two different equations that say when the ball is at a height `2.60` `"m"` and a horizontal range `10.78` `"m"`.

Using the followming equations, let's solve them for `t` to eliminate them:

• `x = x_0 + (v_0cosalpha)t`

`t = (Deltax)/(v_0cosalpha) = color(purple)((10.78)/(v_0(cos38.0^"o"))`

• `y = y_0 + (v_0sinalpha)t - 1/2g t^2`

`t = (v_0sinalpha +-sqrt((-v_0sinalpha)^2 - 4(1/2g)(y-y_0)))/(2(1/2g))`

`= color(green)((v_0sin38.0^"o" +-sqrt((-v_0sin38.0^"o")^2 - 4(4.9)(2.60-2.10)))/(9.8)`

Equating these two equations gives

`color(purple)((10.78)/(v_0(cos38.0^"o")) = color(green)((v_0sin38.0^"o" +-sqrt((-v_0sin38.0^"o")^2 - 4(4.9)(2.60-2.10)))/(9.8)`

After solving for `v_0`, you should get a value of

`color(red)(10.76` `color(red)("m/s"`

Now let's find the maximum speed, by changing te horizontal range to `11.22` `"m"`:

`color(purple)((11.22)/(v_0(cos38.0^"o")) = color(green)((v_0sin38.0^"o" +-sqrt((-v_0sin38.0^"o")^2 - 4(4.9)(2.60-2.10)))/(9.8)`

Solving again for `v_0` gives

`color(blue)(10.96` `color(blue)("m/s"`

Therefore, the range of initial speeds is, with three significant figures,

`v_0 in [color(red)(10.8)color(white)(l)color(red)("m/s"), color(blue)(11.0)color(white)(l)color(blue)("m/s")]`

Answer 2

`u_(min) = 10.76 \ ms^(-1)`
`u_(max)=10.97 \ ms^(-1)`

Explanation:

Modelling the basketball as a point projectile and ignoring air resistance.

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

`{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :}`

Horizontal Motion

The projectile will move under constant speed (NB we can still use "suvat" equation with a=0).

The projectile will travel a distance `S` in time `T`, where `S=11+-0.22m`, ie a minimum distance `s_1=10.78` and a maximum distance `s_2=11.22`

So we can calculate `s` using `s=ut`

`S=ucos(38)T => T=S/(ucos38)`

Vertical Motion

The projectile travels under constant acceleration due to gravity, applied vertically upwards. The projectile must travel a distance `2.6-2.1 = 0.5 \ m` vertically.

`{ (s=,0.5,m),(u=,u sin(38),ms^-1),(v=,"Not Required",ms^-1),(a=,-g,ms^-2),(t=,T=S/(ucos(38)),s) :}`

Applying `s=ut+1/2at^2` we have:

`0.5 = u sin(38) (S)/(ucos(38)) + 1/2(-g)(S/(ucos(38)))^2`

`:. 0.5 = S tan(38) -(gS^2)/(2u^2cos^2(38))`

And re-arranging we get:

`(gS^2)/(2u^2cos^2(38)) = S tan(38) - 0.5`
`u^2 = (gS^2)/(2cos^2(38) (S tan(38) - 0.5))`

If we take `g=9.81`, then:

With `S=S_1=10.78` we get:

`u^2 = 115.867918 ... => u=10.76419 ...`

With `S=S_2=11.22` we get:

`u^2 = 120.29947 ... => u=10.96811 ...`

As we used `g` to `2` decimal places can only expect our solution to be accurate to `2` decimal places, thus we have a rage of `u` from `u_(min) = 10.76` to `u_(max)=10.97`