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Chapter 3: Problem 18

To attain maximum height for the trajectory of a projectile, what angle would you choose between \(0^{\circ}\) and \(90^{\circ}\), assuming that you can launch the projectile with the same initial speed independent of the launch angle. Explain your reasoning.

Short Answer

Expert verified
Answer: To attain the maximum height in the trajectory of a projectile with a constant initial speed, the launch angle should be \(90^{\circ}\).

Step by step solution

01

Understand the projectile motion

A projectile motion is governed by two components: horizontal motion and vertical motion. The horizontal motion has a constant velocity, and the vertical motion is influenced by the acceleration due to gravity. In this problem, we are interested in the vertical motion, as we want to find the maximum height attained by the projectile.
02

Write down the vertical motion equation

To find the relationship between the launch angle and the maximum height, we need to analyze the vertical motion of the projectile. The vertical motion equation is given by: \(h(t) = v_{0y}t - \frac{1}{2}gt^2\), where \(h(t)\) is the height at time \(t\), \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time.
03

Find the initial vertical velocity

The initial vertical velocity can be found using the initial speed of the projectile and the launch angle. It is given by the equation: \(v_{0y} = v_{0}\sin{\theta}\), where \(v_{0}\) is the initial speed, and \(\theta\) is the launch angle.
04

Determine the time to reach maximum height

The projectile reaches its maximum height when its vertical velocity becomes zero. From the vertical motion equation, we can find the time needed to reach the maximum height: \(v_{y}(t) = v_{0y} - gt\), where \(v_y(t)\) is the vertical velocity at time \(t\). Setting \(v_y(t)\) to zero, we can solve for the time \(t_{max}\): \(0 = v_{0}\sin{\theta} - gt_{max}\). Hence, \(t_{max} = \frac{v_{0}\sin{\theta}}{g}\).
05

Find the maximum height

Substituting the values of \(v_{0y}\) and \(t_{max}\) in the height equation (from step 2), we get the maximum height: \(h_{max} = v_{0}\sin{\theta}\frac{v_{0}\sin{\theta}}{g} - \frac{1}{2}g\left(\frac{v_{0}\sin{\theta}}{g}\right)^2\). Simplifying, we get: \(h_{max} = \frac{v_{0}^2\sin^2{\theta}}{2g}\).
06

Determine the angle to maximize height

To find the angle at which maximum height is obtained, we need to maximize the expression for \(h_{max}\) (from step 5) with respect to the angle \(\theta\). Since \(\frac{v_{0}^2}{2g}\) is constant, we need to maximize \(\sin^2{\theta}\). The maximum value for \(\sin^2{\theta}\) is 1, which occurs when \(\theta = 90^{\circ}\). Therefore, to attain maximum height in the trajectory of a projectile with the same initial speed, the launch angle must be \(90^{\circ}\).

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Most popular questions from this chapter

A rocket-powered hockey puck is moving on a (frictionless) horizontal air- hockey table. The \(x\) - and \(y\) -components of its velocity as a function of time are presented in the graphs below. Assuming that at \(t=0\) the puck is at \(\left(x_{0}, y_{0}\right)=(1,2)\) draw a detailed graph of the trajectory \(y(x)\).

In ideal projectile motion, the velocity and acceleration of the projectile at its maximum height are, respectively, a) horizontal, vertical c) zero, zero. downward. d) zero, vertical downward. b) horizontal, zero. e) zero, horizontal.

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