Question

A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from which it is thrown, and if we ignore air resistance, what is the maximum range in meters (which is approximately the same as the range in yards) of a pass at this speed? How long is the ball in the air?

Answer 1

Answer:

The ball was in air for 3.896 s

Explanation:

given,

g = 9.8 m/s², acceleration due to gravity,

If the launch angle is 45°, the horizontal range will be maximum.

The horizontal and vertical launch velocities are equal, and each is equal to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The time to attain maximum height is one half of the time of flight.

v = u + at                     ∵ v = 0 (max. height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The time of flight is twice of the maximum height time

2 t₁ = 3.896 s

The horizontal distance traveled is

D = v × t

D = 3.896×19.09

   = 74.375 m

The ball was in air for 3.896 s