Problem For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height?

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem

For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height?

Solution 

The components of v0 are expressed as follows:

vinitial-x = v0cos(θ)

vinitial-y = v0sin(θ)

 

a)

Let us first find the time it takes for the projectile to reach the maximum height.
Using:  vfinal-y = vinitial-y + ayt

since the y-axis velocity of the projectile at the maximum height is
vfinal-y =___________

Then,

___________= vinitial-y + ayt

Substituting the expression of vinitial-y and ay = -g, results to the following:

__________=__________-__________t

Thus, the time to reach the maximum height is
tmax-height = __________ /__________

We will use this time to the equation

yfinal - yinitial = vinitial-yt + (1/2)ayt2

if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so

hmax = vinitial-yt + (1/2)ayt2

substituting, the vinitial-y expression above, results to the following

hmax =__________t + (1/2)ayt2

Then, substituting the time, results to the following

hmax = (__________x__________/__________) + (1/2)ay(__________/__________)2

simplifying the expression, yields

hmax =__________x sin__________(__________)/__________

b)

The distance traveled by a projectile follows a uniform motion, meaning, velocity is constant from the start point until it reach the ground along the horizontal axis, so, the range R can be expressed as

R = vinitial-xt

Substituting the initial velocity on the x-axis results to the following

R = (__________) t

But, the time it takes a projectile to travel this distance is just twice of tmax-height, by substitution, we obtain the following:

R =__________x 2 x (__________/__________)

Re-arranging and then applying the trigonometric identity

sin(2x) = 2sin(x)cos(x)

we arrive at the expression for the range R as

R =__________sin__________(__________)/__________

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