QUESTION 5 Problem For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 900), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of 0 gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = Vocos(0) Vinitial-y = vosin(e) 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 Thus, the time to reach the maximum height is Imax-height = We will use this time to the equation
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
Substituting ay = -g, results to
hmax = (x/) - (1/2)g(/)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()/
![QUESTION 5
Problem
For a projectile lunched with an initial velocity of vo at an angle of 0 (between 0 and 90°), a) derive the general expression for maximum height hmax and
the horizontal range R. b) For what value of 0 gives the highest maximum height?
Solution
The components of vo are expressed as follows:
Vinitial-x = Vocos(0)
%3D
Vinitial-y = Vosin(0)
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
%3D
Then,
= Vịnitial-y + ayt
Substituting the expression of Minitial-y and ay = -g, results to the following:
%3D
Thus, the time to reach the maximum height is
Imax-height
We will use this time to the equation
Click Save ad Submit to save and subrit. Click Save All Answers to save all answers.
Save All Answers](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13571190-bb6b-4bd5-b20c-9f9b465e0a2f%2Fc9e3d6ea-8ff9-4d96-a2bb-3c0d57f822ae%2F5ty1k8t_processed.jpeg&w=3840&q=75)
![v Question Completion Status:
We will use this time to the equation
Yfinal - Yinitial = Vinitial-yt + (1/2)a,
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)ayt?
%3D
substituting, the Vịnitial-y expression above, results to the following
hmax =
t+ (1/2)ayt
Then, substituting the time, results to the following
hmax (
) + (1/2)ay(
%3D
Substituting ay = -g, results to
hmax = (
) - (1/2)g(
%3D
2
simplifying the expression, yields
hmax =
x sin
%3D
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-x
Substituting the initial velocity on the x-axis results to the following
R= (
But, the time it takes a projectile to travel this distance is just twice of tmax-height by substitution, we obtain the following:
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
Save All Answers
Close W](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13571190-bb6b-4bd5-b20c-9f9b465e0a2f%2Fc9e3d6ea-8ff9-4d96-a2bb-3c0d57f822ae%2F2dbv5ug_processed.jpeg&w=3840&q=75)
![](/static/compass_v2/shared-icons/check-mark.png)
Step by step
Solved in 4 steps with 4 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
![University Physics Volume 1](https://www.bartleby.com/isbn_cover_images/9781938168277/9781938168277_smallCoverImage.gif)
![Glencoe Physics: Principles and Problems, Student…](https://www.bartleby.com/isbn_cover_images/9780078807213/9780078807213_smallCoverImage.gif)
![University Physics Volume 1](https://www.bartleby.com/isbn_cover_images/9781938168277/9781938168277_smallCoverImage.gif)
![Glencoe Physics: Principles and Problems, Student…](https://www.bartleby.com/isbn_cover_images/9780078807213/9780078807213_smallCoverImage.gif)