The key to these projectile questions is to treat the horizontal and vertical components of motion separately and use the fact that they share the same time of flight.
When the baseball has reached its height of 3m it falls under gravity so we can write:
#sf(h=1/2"gt"^(2))#
#:.##sf(t=sqrt((2h)/g))#
#sf(t=sqrt((2xx3.0)/(9.8))=0.782color(white)(x)s)#
This means the total time of flight will be:
#sf(T=2xxt=2xx0.782=1.56color(white)(x)s)#
If #sf(theta)# is the angle of launch then the horizontal component of velocity is constant and is given by:
#sf(V_(x)=Vcostheta)#
Since #sf(V_x=s/T)#
Then:
#sf(V_x=32/1.56=21.51color(white)(x)"m/s")#
So:
#sf(Vcostheta=21.51color(white)(x)"m/s")" "color(red)((1))#
To get #sf(V)# we need to get the angle of launch so now we consider the vertical component of the motion.
We can use the equation of motion:
#sf(v=u+at)#
This becomes:
#sf(0=Vsintheta-"gt")#
#:.##sf(Vsintheta=(9.8xx0.782)=7.66color(white)(x)"m/s"" "color(red)((2)))#
Dividing #sf(color(red)((2)))# by #sf(color(red)((1))rArr)#
#sf((cancel(V)sintheta)/(cancel(V)costheta)=7.66/21.51=0.356=tantheta)#
From which:
#sf(theta=19.6^(@))#
This is the angle of launch.
To get the launch velocity #sf(V)# we can substitute this into #sf(color(red)((1))rArr)#
#sf(Vcos(19.6)=21.51)#
#:.##sf(V=21.51/0.942=22.8color(white)(x)"m/s")#
