Initial Speed of Football Kicked at 45 Degrees & 37m Distance

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A football is kicked at a 45-degree angle, covering a distance of 37 meters and just clearing a 3-meter high goal. To find the initial speed, the discussion emphasizes separating the initial velocity into horizontal and vertical components. The relevant equations of motion are provided, highlighting the need to account for both constant speed and accelerated motion. Participants are encouraged to derive time in terms of the horizontal component and substitute it into the vertical motion equation to solve for the initial velocity. The approach focuses on systematically applying physics principles to arrive at the solution.
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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
 
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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.
 
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.
 
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