Question

the range r and the maximum height h of a projectile fired at an inclination θ to the horizontal with initial speed v₀ are given by the formulas below, where g = 32.2 feet per second per second is the acceleration due to gravity.

r = \frac{2v₀² sinθ cosθ}{g}

h = \frac{v₀² sin²θ}{2g}

a. find the range r if the projectile is fired at an angle of 45° to the horizontal with an initial speed of 150 feet per second.

r ≈ \square feet

(round to two decimal places as needed.)

Explanation:

Step1: Substitute the values into the range formula

Given \(v_0 = 150\) ft/s and \(\theta=45^{\circ}\), \(\sin\theta=\cos\theta=\frac{\sqrt{2}}{2}\), \(g = 32.2\) ft/s².
The range formula is \(R=\frac{2v_{0}^{2}\sin\theta\cos\theta}{g}\).
Substitute the values: \(R=\frac{2\times150^{2}\times\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{2}}{32.2}\).
First, calculate \(150^{2}=22500\), and \(\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{2}=\frac{2}{4}=\frac{1}{2}\).
Then \(2\times22500\times\frac{1}{2}=22500\).
So \(R = \frac{22500}{32.2}\approx700.0\) (rounded to one decimal place as \(\frac{22500}{32.2}\approx700.0\))

Step2: Substitute the values into the height formula

The height formula is \(H=\frac{v_{0}^{2}\sin^{2}\theta}{2g}\).
Substitute \(v_0 = 150\) ft/s, \(\sin\theta=\frac{\sqrt{2}}{2}\), \(g = 32.2\) ft/s².
\(\sin^{2}\theta=\frac{1}{2}\), \(v_{0}^{2}=22500\).
Then \(H=\frac{22500\times\frac{1}{2}}{2\times32.2}=\frac{11250}{64.4}\approx174.7\)

Answer:

\(R\approx700.0\) feet, \(H\approx174.7\) feet