Let `V_0` be the initial speed. Then on the x and y axis we have

`V_(0x) = V_0*cos(alpha)`

`V_(0y) =V_0*sin(alpha)`

On the x axis the motion is uniform

`x = V_(0x)*t_(max) = V_0*t_(max)*cos(alpha)`

On the y axis the motion is uniform accelerated

`0 = V_(0y)^2 -2*g*y`

`y = V_(0y)^2/(2g)`

Time from the maximum height `y` to ground is `t =t_max/2`

`y = g*t^2/2`

Hence

`V_(0y)^2/(2g) =(g*t^2)/2`

`t = (V_(0y)/g) =(V_0/g)*sin(alpha)`

`t_(max) =2*t =(2V_0/g)*sin(alpha)`

Now we get back to the expression of x ` `

`x =2(V_0^2/g)*sin(alpha)*cos(alpha) = (V_0^2/g)*sin(2*alpha)`

Therefore the initial velocity is

`V_0 = sqrt((x*g)/sin(2*alpha)) = sqrt(200*9.81/sin(pi/2)) =44.29 m/s`

In miles per hour this speed is

`V_0 =44.29 m/s =0.04429*3600 (km)/h =`

`=159.46*0.621 mph =99.02 mph`

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