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
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
Thanks for your help
2 Answers
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
Using the followming equations, let's solve them for
#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)(yy_0)))/(2(1/2g))#
#= color(green)((v_0sin38.0^"o" +sqrt((v_0sin38.0^"o")^2  4(4.9)(2.602.10)))/(9.8)#
Equating these two equations gives
After solving for
Now let's find the maximum speed, by changing te horizontal range to
Solving again for
Therefore, the range of initial speeds is, with three significant figures,
# 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:
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
So we can calculate
# 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
# { (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
# 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 rearranging 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
With
# u^2 = 115.867918 ... => u=10.76419 ... #
With
# u^2 = 120.29947 ... => u=10.96811 ... #
As we used