Question

the range r and the maximum height h of a projectile fired at an angle 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.
Since \(\sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}\), and the formula for range \(R=\frac{2v_0^{2}\sin\theta\cos\theta}{g}\).
Substitute the values: \(R=\frac{2\times(150)^{2}\times\sin45^{\circ}\times\cos45^{\circ}}{32.2}\)

Step2: Simplify the trigonometric part

\(\sin45^{\circ}\times\cos45^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{2}=\frac{2}{4}=\frac{1}{2}\)
So the formula becomes \(R=\frac{2\times(150)^{2}\times\frac{1}{2}}{32.2}\)

Step3: Simplify the expression

The \(2\) and \(\frac{1}{2}\) cancel out. Then we have \(R = \frac{(150)^{2}}{32.2}=\frac{22500}{32.2}\)

Step4: Calculate the value

\(R=\frac{22500}{32.2}\approx700.00\) (rounded to two decimal places)

Answer:

\(700.00\)