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^@38.0∘ angle. If the shot is made from a horizontal distance of 11.00 m and must be accurate to +-0.22 m±0.22m (horizontally), what is the range of initial speeds allowed to make the basket?

Thanks for your help

2 Answers
Nathan L.
Jun 30, 2017

v_0 in [10.8color(white)(l)"m/s", 11.0color(white)(l)"m/s"]v0∈[10.8lm/s,11.0lm/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_0y0 is 2.102.10 "m"m

  • final height yy is 2.602.60 "m"m

  • angle of projection alpha_0α0 is 38.0^"o"38.0o

  • 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")]

Steve M
Jul 1, 2017

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)) :}

enter image source here

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

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