Question

1. 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 adjust the angle of the cannon from 20.° to 35.° above horizontal, what effect will she observe on the hang time of the projectiles?

hang time will decrease, but range will increase
hang time will increase, but range will decrease
hang time and range will both increase.
hang time and range will both decrease.

Explanation:

Step1: Determine the formula for hang - time

The formula for the hang - time \(T\) of a projectile is \(T=\frac{2v_{0}\sin\theta}{g}\), where \(v_{0}\) is the initial velocity, \(\theta\) is the launch angle and \(g = 9.8\ m/s^{2}\).
Since \(v_{0}=25\ m/s\) (constant) and \(g\) is constant. When \(\theta\) increases from \(20^{\circ}\) to \(35^{\circ}\), \(\sin\theta\) increases (\(\sin20^{\circ}\approx0.342\), \(\sin35^{\circ}\approx0.574\)). So, \(T\) increases.

Step2: Determine the formula for range

The formula for the range \(R\) of a projectile is \(R=\frac{v_{0}^{2}\sin2\theta}{g}\).
When \(\theta = 20^{\circ}\), \(2\theta=40^{\circ}\) and \(\sin40^{\circ}\approx0.643\). When \(\theta = 35^{\circ}\), \(2\theta = 70^{\circ}\) and \(\sin70^{\circ}\approx0.940\).
Since \(\sin2\theta\) increases (\(0.643<0.940\)), \(R\) increases.

Answer:

Hang time and range will both increase.