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 time - of - flight formula

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

Step2: Analyze time - of - flight change

For $\theta_1 = 45^{\circ}$, $\sin\theta_1=\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.707$. For $\theta_2 = 60^{\circ}$, $\sin\theta_2=\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$. Since $t=\frac{2v_0\sin\theta}{g}$ and $v_0$ and $g$ are constant, as $\sin\theta$ increases from $\sin45^{\circ}$ to $\sin60^{\circ}$, the time in the air $t$ increases.

Step3: Recall range formula

The range formula for a projectile launched from and landing at the same height is $R=\frac{v_0^{2}\sin2\theta}{g}$.

Step4: Analyze range change

For $\theta_1 = 45^{\circ}$, $\sin2\theta_1=\sin90^{\circ}=1$. For $\theta_2 = 60^{\circ}$, $\sin2\theta_2=\sin120^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$. Since $R = \frac{v_0^{2}\sin2\theta}{g}$ and $v_0$ and $g$ are constant, as $\sin2\theta$ decreases from $\sin90^{\circ}$ to $\sin120^{\circ}$, the total horizontal distance $R$ decreases.

Answer:

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