Questions & Answers

Question

Answers

A. 6.25

B. 5.625

C. 11.25

D. 12.5

Answer
Verified

A projectile is fired straight upward with the velocity 90 feet per second at the acceleration due to Earthâ€™s gravity of 32 ft/sec

$v(t)=v(0)+(-32t)$

To find when it hits the ground, find the time when it was at peak and double it.

Its peaks at

$\begin{align}

& v(t)=0 \\

& v(0)=90\text{ (given)} \\

\end{align}$

Place the known values â€“

$\begin{align}

& v(t)=v(0)+(-32t) \\

& 0=90-32t \\

\end{align}$

Take negative terms on the left side of the equation.

$32t=90$

[When the multiplicative term changes its side, it goes in the denominator]

$\begin{align}

& \therefore t=\dfrac{90}{32} \\

& \therefore t=2.8125 \\

\end{align}$

Object hits the ground at time,

$\begin{align}

& t=2.8125\times 2 \\

& t=5.625\text{ Sec} \\

\end{align}$

If you have an initial velocity (if you threw the ball up or down instead of just letting go of it), then you have to include this in the equation, too, giving you: V = Vo + gt, where Vo is the initial velocity of the object. This equation will still work if you threw the ball to the side, instead of straight up or down, except that it will only give you the up-down velocity, not the total velocity. (And the number you should use for Vo is still just the up-down velocity that the object starts with.)

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