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March 6, 2015

March 6, 2015

Posted by **john** on Sunday, December 2, 2007 at 6:53pm.

- calculus -
**drwls**, Sunday, December 2, 2007 at 7:40pmThe equations are:

x = 133 cos(35)* t

y = 5 + 133 sin 35 * t - (1/2) g t^2

It will remain in the air until y = 0. (Solve then y equation for t).

g = 9.8 m/s^2.

Maximum height is attained when

Vy = 133 sin 35 - gt = 0.

Solve for t. That is the time when the height is a maximum. Then insert that t into the equation for y(t). That will give you th3e maximum height.

- calculus -
**john**, Sunday, December 2, 2007 at 8:30pmx = 124..98t; y = 16t^2 + 45.49t + 5

2.729 sec

1.422 sec

4.983 feet

is the answer I'm coming up with ??

- calculus -

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