Question

8 kara is setting up a tee - shirt cannon for a community event. the angle of the cannon can be adjusted, but all projectiles leave the cannon at 25 meters per second. if kara sets the angle of the cannon to be 20.° above horizontal, what will be the range of the projectiles? 64 m 41 m 25 m 61 m

Explanation:

Step1: Recall the range formula for projectile motion

The range formula for projectile motion is \(R=\frac{v^{2}\sin(2\theta)}{g}\), where \(v = 25\space m/s\), \(\theta=20^{\circ}\), and \(g = 9.8\space m/s^{2}\).

Step2: Calculate \(2\theta\)

\(2\theta=2\times20^{\circ} = 40^{\circ}\).

Step3: Find \(\sin(40^{\circ})\)

Using a calculator, \(\sin(40^{\circ})\approx0.6428\).

Step4: Substitute values into the range formula

\(R=\frac{25^{2}\times0.6428}{9.8}=\frac{625\times0.6428}{9.8}=\frac{401.75}{9.8}\approx41\space m\).

Answer:

41 m