B) Find the maximum height ymax. The distance the particle moves upward while slowing with the acceleration −g is ymax = vy0ty max − 1 2g(ty max)2. Therefore, substitute in the expression for ty max from part (a), and simplify, to obtain ymax = v02 sin2 θ 2g. Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. In this case, we have ymax = m. D.Find the range. While in flight, the projectile's horizontal velocity component is constant and x attains the maximum value when t = tflight x = vx0t + 0. Call this maximum value the range R R = v0 cos θ tflight. Substituting the expression for tflight found above and rearranging then gives R = 2v02sin θ cos θ g. Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. Find R. R = m (E) Find the angle that gives the maximum range. To find the maximum range, we will use the trigonometric identity: 2 sin θ cos θ = sin(2θ). Substituting this into the above expression for R gives R = v02 sin(2θ) g. The maximum R occurs when sin(2θ) is at its maximum, which occurs at θmax = °.
B) Find the maximum height ymax. The distance the particle moves upward while slowing with the acceleration −g is ymax = vy0ty max − 1 2g(ty max)2. Therefore, substitute in the expression for ty max from part (a), and simplify, to obtain ymax = v02 sin2 θ 2g. Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. In this case, we have ymax = m. D.Find the range. While in flight, the projectile's horizontal velocity component is constant and x attains the maximum value when t = tflight x = vx0t + 0. Call this maximum value the range R R = v0 cos θ tflight. Substituting the expression for tflight found above and rearranging then gives R = 2v02sin θ cos θ g. Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. Find R. R = m (E) Find the angle that gives the maximum range. To find the maximum range, we will use the trigonometric identity: 2 sin θ cos θ = sin(2θ). Substituting this into the above expression for R gives R = v02 sin(2θ) g. The maximum R occurs when sin(2θ) is at its maximum, which occurs at θmax = °.
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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(B) Find the maximum height ymax.
The distance the particle moves upward while slowing with the acceleration −g is
ymax = vy0ty max −
g(ty max)2.
| 1 |
| 2 |
Therefore, substitute in the expression for ty max from part (a), and simplify, to obtain
ymax =
.
| v02 sin2 θ |
| 2g |
Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. In this case, we have
ymax = m.
D.Find the range.
While in flight, the projectile's horizontal velocity component is constant and x attains the maximum value when t = tflight
x = vx0t + 0.
Call this maximum value the range R
R = v0 cos θ tflight.
Substituting the expression for tflight found above and rearranging then gives
R =
.
| 2v02sin θ cos θ |
| g |
Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. Find R.
R = m
(E) Find the angle that gives the maximum range.
To find the maximum range, we will use the trigonometric identity:
2 sin θ cos θ = sin(2θ).
Substituting this into the above expression for R gives
R =
.
| v02 sin(2θ) |
| g |
The maximum R occurs when sin(2θ) is at its maximum, which occurs at
θmax = °.
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