QUESTION 4 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(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 = 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 tmax-height = We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)ayt 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 Vịnitial-y expression above, results to the following hmax = t+ (1/2)ayt? Then, substituting the time, results to the following
QUESTION 4 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(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 = 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 tmax-height = We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)ayt 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 Vịnitial-y expression above, results to the following hmax = t+ (1/2)ayt? Then, substituting the time, results to the following
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![QUESTION 4
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)
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 =
Then,
= Vịnitial-y + ayt
Substituting the expression of Vinitial-y and ay = -g, results to the following:
%3D
Thus, the time to reach the maximum height is
tmax-height
We will use this time to the equation
Yfinal - Yinitial = Vịnitial-yt +
(1/2)ayt?
%3D
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 vinitial-y expression above, results to the following
hmax =
t+ (1/2)ayt2
Then, substituting the time, results to the following](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c1c9008-8f01-4795-b5d7-6768ba24947f%2F254b0458-2d92-4c24-9b04-e4594117fba2%2F1qg6zs_processed.png&w=3840&q=75)
Transcribed Image Text:QUESTION 4
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)
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 =
Then,
= Vịnitial-y + ayt
Substituting the expression of Vinitial-y and ay = -g, results to the following:
%3D
Thus, the time to reach the maximum height is
tmax-height
We will use this time to the equation
Yfinal - Yinitial = Vịnitial-yt +
(1/2)ayt?
%3D
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 vinitial-y expression above, results to the following
hmax =
t+ (1/2)ayt2
Then, substituting the time, results to the following
![Then, substituting the time, results to the following
hmax = (
)+ (1/2)ay
%3D
Substituting ay = -g, results to
hmax = (
) - (1/2)g(
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-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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c1c9008-8f01-4795-b5d7-6768ba24947f%2F254b0458-2d92-4c24-9b04-e4594117fba2%2Fwekfyrj_processed.png&w=3840&q=75)
Transcribed Image Text:Then, substituting the time, results to the following
hmax = (
)+ (1/2)ay
%3D
Substituting ay = -g, results to
hmax = (
) - (1/2)g(
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-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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