If the pitching wedge the golfer is using gives the ball an initial angle =50, w

Question

If the pitching wedge the golfer is using gives the ball an initial angle =50, what range of velocities v will cause the ball to land within 3 m of the hole? The hole is 30 meters from the golfer (assume the hole lies in the plane of the balls trajectory)

What I got so far in this problem is that time = x/vcos and that y=(vsin)t - (gt2)/2. I substitued the time formula into y. Since the distance from golfer to the hole is 30 meters, x has to be either 27 or 33. When I went about trying to solve this problem I kept getting 13 m/s which is wrong. Could someone please show me where my error is and how to correct it.

Explanation / Answer

First, off the equation should be y = (vsin)t+(gt2)/2

Now, substitue the time equation in for t:

y = (vsin)(x/(vcos))+0.5g(x/(vcos))2

Simplify:

y = xcot + 0.5gx2/v2cos2

Solve for v:

v = ((2cos2(y-xcot))/gx2)

We know:

x = 27 OR x = 33

Solve for y using trig:

y = 27tan55 OR y = 33tan55

Now plug each case into v equation using angles in degrees (NOT radians):

You get the following:

v = 31 m/s OR v = 41.9 m/s

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