Question

1. when a projectile is released, what angle of initial velocity will achieve the greatest range? 15 degrees 30 degrees 45 degrees 60 degrees

Explanation:

Step1: Range formula

The range formula for projectile motion is \(R=\frac{v_{0}^{2}\sin2\theta}{g}\), where \(v_{0}\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

Step2: Maximize \(\sin2\theta\)

We know that the maximum value of the sine function \(y = \sin x\) is \(1\). For \(y=\sin2\theta\), when \(\sin2\theta = 1\), then \(2\theta=90^{\circ}\) (since \(\sin90^{\circ}=1\)).

Step3: Solve for \(\theta\)

If \(2\theta = 90^{\circ}\), then \(\theta = 45^{\circ}\)

Answer:

45 degrees