Let's go through the solutions step by step.

Question 2: Motion of a stone thrown vertically

A stone is thrown upwards and falls back, following the velocity-time graph.

2.1 Acceleration of the stone

Since gravity is the only force acting on the stone (neglecting air resistance), the acceleration is constant and equal to $g = 9.8 \text{ m/s}^2$ downwards.

• 2.1.1 From A to B: The stone is moving upwards but slowing down. The acceleration is $-9.8 \text{ m/s}^2$ (downward).
• 2.1.2 At B: At the peak, the velocity is zero, but acceleration is still $-9.8 \text{ m/s}^2$ (downward).
• 2.1.3 From B to C: The stone is moving downward, increasing in speed. The acceleration remains $-9.8 \text{ m/s}^2$ (downward).

2.2 Value of $t_1$ (time to reach peak)

Using the first equation of motion: $v = u + at$ At the peak (point B), velocity is zero: $0 = u + (-9.8) t_1$ From the graph, initial velocity $u = 7.84 \text{ m/s}$, so:

$t_1 = \frac{7.84}{9.8} = 0.8 \text{ s}$

2.3 Value of $t_2$ (total time for the motion)

The motion is symmetric, meaning the time to go up equals the time to come down: $t_2 = 2 t_1 = 2(0.8) = 1.6 \text{ s}$

2.4 Value of $x_1$ (maximum height)

Using the kinematic equation: $v^2 = u^2 + 2 a s$ Since velocity at peak is zero, we solve for $s$ (height):

$0 = (7.84)^2 + 2(-9.8) s$

$s = \frac{(7.84)^2}{2(9.8)}$

$s = \frac{61.4656}{19.6} = 3.14 \text{ m}$

Question 3: Motion of a cricket ball

A cricket ball is thrown upwards at 14.7 m/s.

3.1 Calculate total time in the air

Using the equation: $t_1 = \frac{u}{g} = \frac{14.7}{9.8} = 1.5 \text{ s}$

Total time: $t_2 = 2 t_1 = 2(1.5) = 3.0 \text{ s}$

3.2 Velocity when returning to the pitcher’s hand

By symmetry, the velocity when it returns to the hand is the same magnitude but opposite direction to the initial velocity.

$v = -14.7 \text{ m/s}$

3.3 Sketch the graphs

You need to draw:

1. Position-Time Graph: A parabolic curve (reaching max height at $1.5$ s, then coming down).
2. Velocity-Time Graph: A straight line decreasing from 14.7 m/s to -14.7 m/s in 3 s.
3. Acceleration-Time Graph: A constant horizontal line at $-9.8 \text{ m/s}^2$.

Would you like any clarifications or more details? Here are some related questions to think about:

1. What happens to acceleration when the stone reaches its peak?
2. How does air resistance affect the motion of the stone?
3. Why is the velocity-time graph a straight line?
4. How would the motion change if the stone was thrown on another planet?
5. Can we calculate the displacement using the area under the velocity-time graph?

Tip: The key to solving motion problems is to identify given values, use the correct equation, and check if motion is symmetrical!