Posted by **Sahil Patel** on Wednesday, September 7, 2011 at 6:42pm.

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?

- algebra -
**Henry**, Saturday, September 10, 2011 at 3:06pm
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.

- algebra -
**Henry**, Saturday, September 10, 2011 at 3:19pm
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.

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