You throw a ball straight into the air from a height of 4 feet and with a speed of 8.22 m/s. The moment the ball leaves your hand you start running away at a speed of 3.36 m/s. How far are you from the ball, the moment it hits the ground?

Answer 1

V = Vo + g*Tr = 0 at max ht.
8.22 - 9.8Tr = 0
9.8Tr = 8.22
Tr = 0.84 s. = Rise time or time to reach max ht.

hmax = ho + Vo*t + 0.5g*t^2
hmax=13.2 + 8.22*0.84 - 4.9*0.84^2=16.7 m. Above gnd.

0.5g*t^2 = 16.7 m.
4.9t^2 = 16.7
t^2 = 3.40
Tf = 1.84 s. = Fall time.

d = 3.36m/s * (0.84+1.84)s = 9.0 m.

Answer 2

Correction: 4 Ft = 1.21 m. Not 13.2 m.
hmax=1.21+8.22*0.84 - 4.9*0.84^2=4.66 m
Above gnd.

0.5g*t^2 = 4.66 m.
4.9t^2 = 4.66
t^2 = 0.95
Tf = 0.97 s. = Fall time.

d = 3.36m/s * (0.84+0.97) = 6.1 m.