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Old Sep 14th 2008, 11:46 AM   #1
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Time, Range, and Height

A stationary ball is kicked so that it moves away at 24 m/s and at an angle of 19 degrees above the ground. Determine the ball's A) hang-time, B) maximum range, & C) maximum height.

Can someone please help me with this?

PB
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Old Sep 14th 2008, 02:16 PM   #2
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I will help you but not do all for you. $\displaystyle a_y(t)=-9.8 m/s^2$.

$\displaystyle v_y(t)=-9.8t+v_{0y}=-9.8t+24\sin 19$ so
$\displaystyle y(t)=-4.9t^2+24\sin 19t$.
While for the x-axis, $\displaystyle a_x(t)=0$.
$\displaystyle v_x(t)=v_{0x}=24\cos 19$ so $\displaystyle x(t)=24\cos 19t$.
Maximum height is when $\displaystyle v_{0y}=0$, just solve it.
Hang-time is when you solve for t when $\displaystyle y(t)=0$.
And maximum range is when $\displaystyle y(t)=0$, solve for $\displaystyle t$ and substitute this $\displaystyle t$ into $\displaystyle x(t)$.
And don't forget the units.
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