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Chapter 3: Problem 59

A projectile is launched at \(t=0\) with initial speed \(v_{\mathrm{i}}\) at an angle \(\theta\) above the horizontal. (a) What are \(v_{x}\) and \(v_{y}\) at the projectile's highest point? (b) Find the time \(t\) at which the projectile reaches its maximum height. (c) Show that the maximum height \(H\) of the projectile is $$H=\frac{\left(v_{i} \sin \theta\right)^{2}}{2 g}$$

Short Answer

Expert verified
Show that the maximum height of the projectile is represented by the formula $$H = \frac{(\left(v_i \sin(\theta)\right))^2}{2g}$$.

Step by step solution

01

Find the horizontal and vertical velocities at the highest point

To find the velocities at the highest point, we can use the following kinematic equations: $$v_x = v_i \cos(\theta)$$ $$v_y = v_i \sin(\theta) - gt$$ At the highest point, the vertical velocity \(v_y = 0\). Hence, $$0 = v_i \sin(\theta) - gt$$ Now, we can solve for the horizontal velocity \(v_x\): $$v_x = v_i \cos(\theta)$$
02

Find the time when the projectile reaches the maximum height

To find the time \(t\) at which the projectile reaches its maximum height, we can use the equation for vertical velocity \(v_y = 0\). We already have, $$0 = v_i \sin(\theta) - gt$$ Now, we can solve for the time \(t\): $$t = \frac{v_i \sin(\theta)}{g}$$
03

Show that the given formula represents the maximum height of the projectile

We can find the maximum height \(H\) using the following equation: $$H = v_y t - \frac{1}{2}gt^2$$ Using the values we found from Steps 1 and 2, $$H = 0 \cdot \frac{v_i \sin(\theta)}{g} - \frac{1}{2}g\left(\frac{v_i \sin(\theta)}{g}\right)^2$$ Simplifying the equation, we get, $$H = \frac{(\left(v_i \sin(\theta)\right))^2}{2g}$$ Thus, we have proved that the maximum height \(H\) of the projectile is, $$H = \frac{(\left(v_i \sin(\theta)\right))^2}{2g}$$

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