# Question #46750

##### 1 Answer

There is no single solution. One solution is

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:

Since it hasn't been specified, we will assume the ball is being thrown from the ground,

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=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