for a projectile lunched with an initial velocity of v0 at an angle of between 0 and 90o a deriv

a) The maximum height (hmax) and the horizontal range (R) of a projectile can be derived from th

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? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) 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: = - 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)ayt2 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)ayt2 Then, substituting the time, results to the following hmax = (x/) + (1/2)ay(/)2 Substituting ay = -g, results to hmax = (x/) - (1/2)g(/)2 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()/

          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?
Solution 
The components of v0 are expressed as follows:
vinitial-x = v0cos(θ)
vinitial-y = v0sin(θ)
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:
 =  - 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)ayt2
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)ayt2
Then, substituting the time, results to the following
hmax = (x/) + (1/2)ay(/)2
Substituting ay = -g, results to
hmax = (x/) - (1/2)g(/)2
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()/

Added by Michelle T.

Question

Please give Ace some feedback