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_____(_____)/______
Problem
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_____(_____)/______
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