Physics

Steven W. Walker

September 6, 2011

1 Finding Horizontal Displacement

A projectile launched at angle θ to the horizontal reaches maximum heighth. Show that its horizontal range is:

4h

tan(θ)

Find the Time

The first step to solving this problem is to find the time travelled. To dothis, use the equation:

vf = vi + a ∗ t

As we know a particle has a velocity of 0 at it’s maximum height, vf shouldequal 0. vi should correspond to the initial velocity in the vertical vector,or vi ∗ sin θ. Acceleration (a) should be equal to the force of gravity, or −g

(recall the force is in the downward direction. We need to solve this equationfor t which should be:

t =vi ∗ sin θ

g

Finding Horizontal Displacement

Now that we know how long the particle travelled, we can use that timeto find the displacement in the horizontal vector. This can be accomplishedusing

x = vi ∗ cos θ ∗ 2t

It is important to note the t changed to 2t as we defined t as the time toreach the maximum height. The time to come back down should double the

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value of t for total flight time. Now, replace t with the value we found sothat:

x =vi ∗ cos θ ∗ 2 ∗ vi ∗ sin θ

g

Which simplifies to:

x =2v2i ∗ cos θ sin θ

g

Rearrangement

Now this would be the end of the problem if your professor wasn’t a bastardwe didn’t have to rearrange the equation to the proper form. Let’s manipulatethe formula

v2f = v2i + 2ay

where v2f = 0 as we are again considering time when the particle reachesthe maximum height, h. y2i again refers to the initial velocity in the verticalvector, only squared this time, thus it could be written v2i sin

2 θ. a shouldonce again be replaced by −g and y, representing the vertical displacement,should be replaced with h. The resulting equation is:

0 = v2i sin2 θ − 2gh

We should then solve this for v2i . There is no good reason for this otherthan we need to make the equation we solved in part II look like the answer.Solved for v2i , the equation should be

v2i =2gh

sin2 θ

Now this should be plugged back into the solution to the entire problem wegot earlier so that it replaces v2i . The solution should then read:

x =4gh ∗ cos θ sin θ

g ∗ sin2 θ

The two gs should cancel out. Additionally recall that:

cos θ sin θ

sin2 θ=

cos θ

sin θ=

1

tan θ

Making this replacement, we yield the final solution:

4h

tan θ

Q.E.D. and Happy Math-ing!

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