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

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The discussion focuses on calculating the initial speed of a football kicked at a 45-degree angle, which travels a horizontal distance of 37 meters and just clears a 3-meter high goal. The relevant kinematic equations include v = v_0 + a*t and v^2 = (v_0)^2 + 2a*(x - x_0). Participants emphasized the need to separate the initial velocity into horizontal (v_ox) and vertical (v_oy) components and correctly apply the equations of motion to derive the initial speed. The approach involves substituting known quantities into the vertical motion equation to solve for v_0.

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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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