A basketball player stands 15m away from the basket to make a 3-point shot. The basket is 3m (10 ft) above the ground. Assume the ball leaves the player's hands at a height of 2m, and with velocity vector Vo=(Vox, Voy)?

a) what is the minimum initial y-direction velocity (Voy) needed for the ball to just barely reach the height of the basket?
b) in the scenario of part a), say the ball has Voy initial y-direction velocity. What should Vox be in order for the player to make the shot? (give your answer in terms of general Voy before plugging in the numbers!)
c) Would Voy from part a) be larger or smaller if the player was throwing the ball on the Moon? why?

1 Answer
May 11, 2018

I got this; however I am not sure I interpreted the question correctly...

Explanation:

Consider the diagram:

enter image source here

where:
#H=3m#
#h=2m#
#d=15m#

Let us divide our problem into a component along the #x# axis and #y# axis; along the #y# axis we have the acceleration of gravity #g# operating on the ball while horizontally no acceleration will interfere (assuming negligible air restance).

Let us start considering the initial vertical component of #v_o# needed to reach the height of the basket along #y# (part a, distance #s#):

we can consider the expression from Kinematics:

#v_f^2=v_i^2+2as#

with:

#v_f=v_B=0# ( barely reach the height of the basket )
#v_i=v_(0y)#
#s=y_f-y_i=3m-2m=1m#
#a=-9.8m/s^2#

we get:

#0=v_(0y)^2-2*9.8*1#

so that: #v_(0y)=sqrt(19.6)=4.4m/s#

Let us now use the relationship from Kinematics:

#v_f=v_i+at#

#0=v_(0y)-9.8t#

#t=v_(0y)/9.8=4.4/9.8=0.45s# to go from A to B.

#------------#

Let us now consider part b; to get #v_(0x)# we know that #v_(0y)=4.4m/s# and the ball has to travel #15m# horizontally to get to the basket.

Along #x# there is no acceleration so we have:

#v_(0x)=d/t#

where #t# is the time of flight that must be equal to the vertical time of flight from A to B!
So we end up with:

#v_(0x)=d/(v_(0y)/9.8)=15/0.45=33.3m/s#

#------------#

On the Moon #g# wold be smaller (#g_("Moon")=1.6m/s^2#) so that we get (part a):

#0=v_(0y)^2-2*1.6*1#

so that: #v_(0y"Moon")=sqrt(3.2)=1.8m/s#