Problem 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(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 tmax-height = We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)ay? 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)ayı? Then, substituting the time, results to the following hmax = ( )+ (1/2)ay( Substituting ay = -g, results to hmax = ( )- (1/2)g( 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
Problem 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(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 tmax-height = We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)ay? 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)ayı? Then, substituting the time, results to the following hmax = ( )+ (1/2)ay( Substituting ay = -g, results to hmax = ( )- (1/2)g( 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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Transcribed Image Text: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 e 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
%3D
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)ay?
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(
Substituting ay = -g, results to
hmax = (
)- (1/2)g(
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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