Question

6 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. for safety reasons, event organizers have requested all tee - shirts be launched at an angle greater than 50.° given this constraint, what is the maximum range kara can achieve with the tee - shirt cannon 68 m 46 m 51 m 63 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 \(v = 25\ m/s\) (initial velocity), \(g = 9.8\ m/s^{2}\) (acceleration due to gravity).

Step2: Analyze the sine function for the given angle constraint

We know that \(y = \sin(x)\) is a periodic function. Given \(\theta>50^{\circ}\), then \(2\theta>100^{\circ}\). The function \(y=\sin(x)\) is decreasing in the interval \((90^{\circ}, 270^{\circ})\). The maximum value of \(\sin(2\theta)\) for \(\theta>50^{\circ}\) occurs when \(2\theta = 100^{\circ}\) (since we want the largest possible value of \(\sin(2\theta)\) given the constraint \(\theta>50^{\circ}\)), and \(\sin(100^{\circ})\approx0.985\).

Step3: Substitute the values into the range formula

Substitute \(v = 25\ m/s\), \(g=9.8\ m/s^{2}\) and \(\sin(2\theta)=\sin(100^{\circ})\approx0.985\) into \(R=\frac{v^{2}\sin(2\theta)}{g}\).

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Answer:

63 m