Question

1. a toy rocket is launched twice into the air from level ground and returns to level ground. the rocket is first launched with initial speed v at an angle of 45° above the horizontal. it is launched the second time with the same initial speed, but with the launch angle increased to 60° above the horizontal. describe how both the total horizontal distance the rocket travels and the time in the air are affected by the increase in launch angle. neglect friction.

Explanation:

Step1: Recall projectile - motion formulas

The time - of - flight formula for a projectile launched from and landing at the same height is $t=\frac{2v_0\sin\theta}{g}$, and the horizontal range formula is $R = \frac{v_0^{2}\sin2\theta}{g}$, where $v_0$ is the initial speed, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity.

Step2: Analyze the time - of - flight

For the first launch with $\theta_1 = 45^{\circ}$, $t_1=\frac{2v\sin45^{\circ}}{g}=\frac{2v\times\frac{\sqrt{2}}{2}}{g}=\frac{\sqrt{2}v}{g}$. For the second launch with $\theta_2 = 60^{\circ}$, $t_2=\frac{2v\sin60^{\circ}}{g}=\frac{2v\times\frac{\sqrt{3}}{2}}{g}=\frac{\sqrt{3}v}{g}$. Since $\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$ and $\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.707$, and $t\propto\sin\theta$, the time in the air increases as the launch angle increases.

Step3: Analyze the horizontal range

For $\theta_1 = 45^{\circ}$, $\sin2\theta_1=\sin90^{\circ}=1$, so $R_1=\frac{v^{2}\sin90^{\circ}}{g}=\frac{v^{2}}{g}$. For $\theta_2 = 60^{\circ}$, $\sin2\theta_2=\sin120^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$. Since $R\propto\sin2\theta$ and $\sin120^{\circ}<\sin90^{\circ}$, the horizontal range decreases as the launch angle is increased from $45^{\circ}$ to $60^{\circ}$.

Answer:

The time in the air increases and the horizontal range decreases when the launch angle is increased from $45^{\circ}$ to $60^{\circ}$.