Football kicked at 45° clears 3m goal at 37m distance

[Quevvy]

1. A football is kicked at an angle of 45 degrees from a distance of 37 meters and just clears a 3 meter high goal on its way down. To the nearest m/s what was its initial speed?



2. v = v_0 + a*t
v^2 = (v_0)^2 + 2a*(x - x_0)
x=x_0 + (v_0)t + (1/2)*a*t^2



3. I set the original velocity as V and tried to solve but I got a negative number

 

[Delphi51]

Separate the initial velocity v into horizontal and vertical parts. Then write two headings:
Horizontal and Vertical. Decide in each case whether you have constant speed or accelerated motion and write the appropriate formula(s). Your x = formula is not correct. You don't have a y = formula. Put in the numbers or expressions for all known quantities and try to make progress on all three fronts.

 

[rl.bhat]

First of find the vertical and horizontal components of v_o.
Displacement is given, . Find time t in terms of v_ox.
In the last equastion
y = y_o + v_oy*t - 1/2*g*t^2, substitute the value of t and solve for v_o.