Question

A bad punter on a football team kicks a football approximately straight upward with an initial velocity of 89 ft/sec.

a. If the ball leaves his foot from a height of 4 ft, write an equation for the vertical height s (in ft) of the ball t seconds after being kicked.

b. Find the time(s) at which the ball is at a height of 102.2125 ft. Round to 1 decimal place.

Answer 1

Answer:

See below

Step-by-step explanation:

Vertical velocity will be affected by gravity in this scenario

df = do + vo t + 1/2 a t^2         do = original height = 4 ft      a = -32.2 ft/s^2

df = 4 + 89 t  - 1/2 (32.2) t^2        df = height

On the way up and the way down, the ball may reach height of 102.2125 ft :

102.2125 = 4 + 89 t - 1/2 (32.2) t2   re-arrange to:

-16.1 t^2 + 89t - 98.2125 =0  

      Use Quadratic Formula to find t =   1.5 and 4.0 s