A baseball player releases a baseball at the height of 7 feet above the ground with an initial vertical velocity of 54 feet per second. How long will it take the ball to reach the ground?

Answer 1

Vf^2 = Vo^2 + 2gd,
d = (Vf^2 - Vo^2) / 2g,
d = 7 + (0 - (54)^2) / -64 = 52.56 Ft.
= Distance above ground.

Vf = Vo + gt,
t(up) = (Vf - Vo) / g,
t(up) = (0 - 54) / -32 = 1.69s.

d = Vo*t + 0.5gt^2 = 52.56 Ft.
0 + 0.5*32t = 52.56,
16t = 52.56,
t(down) = 3.29s.

T = t(up) + t(down) = 1.69 + 3.29 = 4.98s = Time to reach ground.

Answer 2

Correction:
0 + 0.5*32t^2 = 52.56,
16t^2 = 52.56,
t^2 = 3.29,
t(down) = 1.81s.

T = t(up) + t(down) = 1.69 + 1.81 = 3.50s. = Time to reach ground.

Answer 3

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