The height of an object above the ground at time(adsbygoogle = window.adsbygoogle || []).push({}); tis given by

s = v_{0}t - (g/2)t^{2}

where v_{0}is the initial velocity and g is the acceleration due to gravity.

(a) At what height is the object initially?

(b) How long is the object in the air before it hits the ground?

(c) When will the object reach its maximum height?

(d) What is that maximum height?

Here I go...

(a)

Just plugged in a zero for time, and solved for s to see the position of the object at time 0 (initial time).

s = v_{0}t - (g/2)t^{2}

s = v_{0}(0) - (g/2)(0)^{2}

s = 0

It's initially on the ground. (at height 0.)

(b)

I knew that the position had to be s = 0 for it to hit the ground, and I know it started from the ground, so one t-intercept would be the starting time, and the other would be the ending time, so i solved the equation for t, using s = 0.

0 = v_{0}t - (g/2)t^{2}

0 = t(v_{0}- (g/2)t)

t = 0 <--- the starting time, and i'll just worry about the other root..

v_{0}- (g/2)t = 0

v_{0}= (g/2)t

(2/g)v_{0}= t

t = (2v_{0})/g

The object is in the air (2v_{0})/g before it hits the ground.

(c)

For this one, since I know it's a parabola, I know that it will have a maximum where the t-coordinate will be the time of the highest point, and the s-coordinate will be the highest point. I knew that to find the vertex, the formula is -b/(2a).

s = (-g/2)t^{2}+ v_{0}t

Vertex = -b/(2a)

-v_{0}/ (2*(-g/2))

-v_{0}/ -g

v_{0}/g

The object will reach its maximum height at time v_{0}/g

(d)

Since I already had the time, I just plugged in the time to get the actual height, s.

s = v_{0}(v_{0}/g) - (g/2)(v_{0}/g)^{2}

s = v_{0}^{2}/g - v_{0}^{2}/(2g)

s = (2v_{0}^{2})/(2g) - v_{0}^{2}/(2g)

s = (2v_{0}^{2}- v_{0}) / (2g)

s = (v_{0}(2v_{0}-1)) / (2g) <--- maximum height

It was a pretty lengthy problem, and there weren't really any numbers, so I just wanted to make sure that everything was correct, thanks.

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# Position of object above the ground.

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