Question

the range r and the maximum height h of a projectile fired at an inclination θ to the horizontal with initial speed ( v_0 ) are given by the formulas below, where ( gapprox32.2 ) feet per second per second is the acceleration due to gravity.
( r=\frac{2v_0^2sin\thetacos\theta}{g} )
( h=\frac{v_0^2sin^2\theta}{2g} )
complete parts a and b.
a. find the range r if the projectile is fired at an angle of ( 45^{circ} ) to the horizontal with an initial speed of 150 feet per second.
( rapproxsquare ) feet
(round to two decimal places as needed.)

Explanation:

Step1: Substitute the values into the range formula

Given \(v_0 = 150\) feet per second, \(\theta=45^{\circ}\), and \(g = 32.2\) feet per second².
The formula for the range \(R=\frac{2v_0^{2}\sin\theta\cos\theta}{g}\).
Since \(\sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}\), then \(\sin\theta\cos\theta=\frac{1}{2}\sin(2\theta)\) (using the double - angle formula \(\sin(2\alpha)=2\sin\alpha\cos\alpha\)). When \(\theta = 45^{\circ}\), \(\sin(2\theta)=\sin(90^{\circ}) = 1\).
Substitute \(v_0 = 150\), \(g = 32.2\) into \(R=\frac{2v_0^{2}\sin\theta\cos\theta}{g}\), we get \(R=\frac{2\times150^{2}\times\frac{1}{2}}{32.2}\).

Step2: Calculate the value of \(R\)

First, simplify the numerator: \(2\times150^{2}\times\frac{1}{2}=150^{2}=22500\).
Then, \(R=\frac{22500}{32.2}\approx700.00\) (rounded to two decimal places).

Answer:

\(700.00\)