Question #46750

1 Answer
Aug 24, 2014

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