To find the time $t_R$ at which the projectile hits the ground, we need to analyze the vertical motion of the projectile using kinematic equations.

Step 1: Define known variables

• $v_0$: initial speed of the projectile
• $\theta$: angle at which the projectile is fired with respect to the horizontal
• $g$: acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$)
• $t$: time since the projectile was fired

Step 2: Vertical motion equation

The vertical position $y(t)$ of the projectile as a function of time is given by the following kinematic equation:

$y(t) = v_{0y} t - \frac{1}{2} g t^2$

Where:

• $v_{0y} = v_0 \sin \theta$ is the initial vertical velocity component.

Since the projectile is fired from and lands back on the ground, we set $y(t_R) = 0$ to find the time of flight $t_R$, which is when the projectile hits the ground.

Step 3: Solve for $t_R$

Substitute $y(t_R) = 0$ and $v_{0y} = v_0 \sin \theta$ into the vertical motion equation:

$0 = v_0 \sin \theta \, t_R - \frac{1}{2} g t_R^2$

This simplifies to:

$t_R \left( v_0 \sin \theta - \frac{1}{2} g t_R \right) = 0$

One solution is $t_R = 0$, which corresponds to the initial launch time. The second solution comes from solving:

$v_0 \sin \theta - \frac{1}{2} g t_R = 0$

$v_0 \sin \theta = \frac{1}{2} g t_R$

$t_R = \frac{2 v_0 \sin \theta}{g}$

Final Answer:

The time $t_R$ at which the projectile hits the ground is:

$t_R = \frac{2 v_0 \sin \theta}{g}$

This is the total time of flight.

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Would you like further details or have any specific questions?

Here are 5 related questions to explore this topic:

1. How do you calculate the maximum height of the projectile?
2. What is the horizontal range of the projectile?
3. How does changing the launch angle affect the time of flight?
4. What would happen if the projectile were launched from an elevated platform?
5. How do air resistance and drag affect projectile motion?

Tip: For maximum range, the ideal launch angle is $45^\circ$.