Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
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?

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?

icon
Related questions
Question

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?

 

Please see images for full problem.

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)ayl 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= ( 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
Transcribed Image Text: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)ayl 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= ( 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)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)ay? substituting, the Vinitial-y expression above, results to the following hmax = ++(1/2)ay?
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)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)ay? substituting, the Vinitial-y expression above, results to the following hmax = ++(1/2)ay?
Expert Solution
steps

Step by step

Solved in 6 steps with 17 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer