For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of e gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = vocos(e) Vinitial-y = vosin(0) a) Let us first find the time 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)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)ayt? substituting, the vinitial-y expression above, results to the following hmax = t+ (1/2)ay? Then, substituting the time, results to the following hmax = ( )+ (1/2)ay(
For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of e gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = vocos(e) Vinitial-y = vosin(0) a) Let us first find the time 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)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)ayt? substituting, the vinitial-y expression above, results to the following hmax = t+ (1/2)ay? Then, substituting the time, results to the following hmax = ( )+ (1/2)ay(
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![Substituting ay = -g, results to
hmax = (
)- (1/2)g(
simplifying the expression, yields
hmax =
x sin
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 Ras
R =
sin](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F91108a7c-b9c9-4ea6-8772-42b37666b4b1%2Fc0f6573e-c281-454f-858a-a8097a59dac0%2F1eim2dr_processed.png&w=3840&q=75)
Transcribed Image Text:Substituting ay = -g, results to
hmax = (
)- (1/2)g(
simplifying the expression, yields
hmax =
x sin
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 Ras
R =
sin
![For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal
range R. b) For what value of e gives the highest maximum height?
Solution
The components of vo are expressed as follows:
Vinitial-x = vocos(e)
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
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 vinitial-y expression above, results to the following
hmax =
t+ (1/2)ayt?
Then, substituting the time, results to the following
hmax = (
)+ (1/2)ay](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F91108a7c-b9c9-4ea6-8772-42b37666b4b1%2Fc0f6573e-c281-454f-858a-a8097a59dac0%2Fu5b33d7_processed.png&w=3840&q=75)
Transcribed Image Text:For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal
range R. b) For what value of e gives the highest maximum height?
Solution
The components of vo are expressed as follows:
Vinitial-x = vocos(e)
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
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 vinitial-y expression above, results to the following
hmax =
t+ (1/2)ayt?
Then, substituting the time, results to the following
hmax = (
)+ (1/2)ay
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