For a projectile lunched with an initial velocity of vo at an angle of 8 (between O 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?

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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(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)ayt?
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
Substituting ay - -g, results to
hmax = (
)- (1/2)g(
simplifying the expression, yields
hmax =
x sin
Transcribed Image Text: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(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)ayt? 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 Substituting ay - -g, results to hmax = ( )- (1/2)g( simplifying the expression, yields hmax = x sin
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