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Kindly fill in the blanks. 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?   The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ)

Kindly fill in the blanks. 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?   The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ)

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College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Kindly fill in the blanks.

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?

 

The components of v0 are expressed as follows:

vinitial-x = v0cos(θ)

vinitial-y = v0sin(θ)

 

 

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 = 0 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)ayt2 substituting, the Vinitial-y expression above, results to the following hmax = 1+ (1/2)ayı?
Transcribed Image Text: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 = 0 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)ayt2 substituting, the Vinitial-y expression above, results to the following hmax = 1+ (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
Transcribed Image Text: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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