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}

complete parts a and b.

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 ≈ 698.76 feet

(round to two decimal places as needed.)

b. find the maximum height h if the projectile is fired at an angle of 45° to the horizontal with an initial speed of 150 feet per second.

h ≈ 74.7 feet

(round to two decimal places as needed.)

Explanation:

Step1: Substitute values for part A

Given \(v_0 = 150\) feet per second, \(\theta=45^{\circ}\), \(g = 32.2\) feet per second².
For the range formula \(R=\frac{2v_0^{2}\sin\theta\cos\theta}{g}\), since \(\sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}\), then \(\sin\theta\cos\theta=\frac{1}{2}\sin(2\theta)\) (double - angle formula \(\sin(2\alpha)=2\sin\alpha\cos\alpha\)). Here \(\theta = 45^{\circ}\), so \(\sin\theta\cos\theta=\frac{1}{2}\).
Substitute into \(R\) formula: \(R=\frac{2\times150^{2}\times\frac{1}{2}}{32.2}\)

$$R=\frac{150^{2}}{32.2}=\frac{22500}{32.2}\approx698.76$$

Step2: Substitute values for part B

For the maximum - height formula \(H=\frac{v_0^{2}\sin^{2}\theta}{2g}\).
Since \(\theta = 45^{\circ}\), \(\sin\theta=\frac{\sqrt{2}}{2}\), \(\sin^{2}\theta=\frac{1}{2}\), \(v_0 = 150\) feet per second, \(g = 32.2\) feet per second².
Substitute into \(H\) formula: \(H=\frac{150^{2}\times\frac{1}{2}}{2\times32.2}=\frac{22500}{4\times32.2}=\frac{22500}{128.8}\approx174.7\)

Answer:

A. \(R\approx698.76\) feet
B. \(H\approx174.7\) feet