Posted by **BJ** on Sunday, October 16, 2011 at 7:54am.

ball is thrown up at 88 ft/s from a height of 25 feet. Path of ball is

y = -16t^2 + 88t +25. y is the height of ball above the ground t seconds after being thrown. when does it reach maxiumum height? (I got that at 2.75 s)

At what time does the ball hit the ground? (I set equation to =0, but I got 5.2 s and answer is 5.77 s)

- math -
**tchrwill**, Sunday, October 16, 2011 at 10:56am
Vf = Vo(t) - 32(t)

Vf = 88 - 32t or t = 2.75sec.

Height reached from 25 ft. height

h = 88(t) - 32(t^2)/2

h = 88(2.75) - 16(2.75)^2 = 121 ft.

It takes the same 2.75 sec. to return to the original launch height of 25 feet and an increment more to fall the 25 feet further to the ground.

Therefore,

h = Vo(t) + 32(t^2)/2

(121 + 25) = 0(t) + 16(t^2)

t - sqrt(146/16) = 3.02

The time from launch to impact is therefore (2.75 + 3.02) = 5.77 sec.

## Answer This Question

## Related Questions

- math - a baseball is hit with an initial velocity of 75 feet per second from the...
- Math - If you are standing near the edge of the top of a 200 feet building and ...
- Algebra - The height of a ball thrown directly up with a velocity of 40 feet per...
- Pre-Calc - ball thrown vertically upward with an initial velocity of 80 ft per ...
- math - An object that is falling or vertically projected into the air has it ...
- Algebra - If a ball is thrown vertically upward from a height of 56ft. above ...
- Math - If you are standing near the edge of the top of a 200 feet building and ...
- Math - If you are standing near the edge of the top of a 200 feet building and ...
- pre-calc - Physics: a ball is thrown vertically upward with an initial velocity ...
- Math- - A ball is thrown vertically upward with an initial velocity of 96 feet ...

More Related Questions