Question #ae3f1
1 Answer
Explanation:
We're asked to find the speed at which the ball was hit, given its horizontal range and hit angle.
To do this, we can use the equations
#ul(Deltay = v_0sinalpha_0t  1/2g t^2#
#ul(Deltax = v_0cosalpha_0t#
where

#Deltay# is the change in height (#0# since the ball landed at the same height it was hit) 
#Deltax# is the change in horizontal position, given as#103# #"m"# 
#v_0# is the initial speed (what we're trying to find 
#alpha_0# is the hit angle, given as#50^"o"#
Plugging in known values to both equations:
#0 = v_0sin(50^"o")t  1/2(9.81color(white)(l)"m/s"^2)t^2#
# 103color(white)(l)"m" = v_0cos(50^"o")t#
We can realize that the time
#t = (103)/(v_0(0.643)) = 160.2/(v_0)#
#0 = v_0(160.2/(v_0))  1/2(9.81)(160.2/(v_0))^2#
Now we solve for
#4.905(160.2/(v_0))^2 = 160.2#
#125944/((v_0)^2) = 160.2#
#v_0 = sqrt((125944)/(160.2)) = color(red)(ulbar(stackrel(" ")(" "28.0color(white)(l)"m/s"" "))#