A quarterback is asked to throw a football to a receiver that is 32.6 m away. What is the minimum speed that the football must have when it leaves the quarterback's hand? Ignore air resistance. Assume the ball is caught at the same height as it is thrown.

Answer 1

The formula you need to use for the distance R of the pass, which is easily derived, is

R = 32.6 = (Vo^2/g)*sin(2A)

Vo is the velocity at which the ball is thrown

For the minimum required speed, make A = 45 degrees, so that 2A = 90 degrees.

32.6 = Vo^2/g
Vo = 17.9 m/s

Answer 2

L=vₒ²•sin2α/g,
vₒ =sqrt(L•g/sin2α)
min vₒ at sin2α = 1
vₒ =sqrt(L•g) = sqrt(32.6•9.8) =17.87 m/s