Question

which projectile will reach a larger maximum height - a projectile launched at a 41° angle or the same projectile launched at a 48° angle?
same height for both
a projectile launched at a 48° angle
a projectile launched at a 41° angle

Explanation:

Step1: Recall the formula for maximum height of a projectile

The formula for the maximum height \(H\) of a projectile is \(H=\frac{v_{0}^{2}\sin^{2}\theta}{2g}\), where \(v_{0}\) is the initial velocity, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity. Since \(v_{0}\) and \(g\) are the same for both projectiles (same projectile, same launch - environment), we only need to compare \(\sin^{2}\theta\) values.

Step2: Calculate \(\sin^{2}\theta\) for each angle

For \(\theta = 41^{\circ}\), \(\sin(41^{\circ})\approx0.656\), and \(\sin^{2}(41^{\circ})\approx0.656^{2}=0.430\).
For \(\theta = 48^{\circ}\), \(\sin(48^{\circ})\approx0.743\), and \(\sin^{2}(48^{\circ})\approx0.743^{2}=0.552\).

Answer:

B. A projectile launched at a \(48^{\circ}\) angle