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?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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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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)
Transcribed Image Text: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)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 = (
) + (1/2)ay(
Substituting ay = -g, results to
hmax = (
) - (1/2)g(
simplifying the expression, yields
hmax =
x sin
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