At t = 0 a rock is projected from ground level with a speed of 15.0 m/s and at an angle of 53.0∘ above the horizontal. Neglect air resistance. At what two times t1 and t2 is the rock 5.00 m above the ground? (Suppose that t1<t2). Express your answers in seconds separated by a comma.

To solve this problem, we will use the kinematic equation for vertical motion, considering that the initial vertical position is at ground level and air resistance is neglected.

The vertical position \( y(t) \) as a function of time \( t \) is given by the equation:

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

Where:
- \( y(t) \) is the height above the ground at time \( t \),
- \( v_{0y} = v_0 \sin(\theta) \) is the initial vertical component of velocity,
- \( v_0 = 15.0 \, \text{m/s} \) is the initial speed,
- \( \theta = 53.0^\circ \) is the angle of projection,
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

We want to find the times \( t_1 \) and \( t_2 \) when the rock is 5.00 meters above the ground, so \( y(t) = 5.00 \, \text{m} \). First, let's calculate \( v_{0y} \):

\[
v_{0y} = v_0 \sin(\theta) = 15.0 \times \sin(53.0^\circ)
\]

Now, substituting the values into the kinematic equation:

\[
5.00 = v_{0y} t - \frac{1}{2} g t^2
\]

We'll solve this quadratic equation for \( t \). Let me calculate it step by step.The two times when the rock is 5.00 meters above the ground are:

\[
t_1 = 0.534 \, \text{seconds}, \quad t_2 = 1.911 \, \text{seconds}
\]

If you have any further questions or need additional details, feel free to ask!

Here are 5 related questions:
1. How high will the rock go at its peak?
2. What is the total time of flight before the rock hits the ground?
3. What is the horizontal distance traveled by the rock when it reaches 5 meters?
4. What is the rock's speed at 5 meters above the ground?
5. What are the horizontal and vertical velocity components at time \( t_1 \)?

**Tip**: When solving kinematic problems, it's useful to break velocity into horizontal and vertical components for easier analysis of motion.

Solve a kinematics problem involving a rock projected at 15 m/s and an angle of 53° above the horizontal. Discover how to calculate the two times the rock is 5 meters above the ground using kinematic equations and the quadratic formula. Suitable for Grades 10-12.

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