To solve this problem, let's break it down step-by-step using the equations of motion under constant acceleration due to gravity. We'll assume $g = 9.8 \, \text{m/s}^2$.

(a) Maximum Height

To find the maximum height reached by the ball, we use the following kinematic equation:

$v^2 = u^2 + 2a s$

Where:

• $v$ is the final velocity at the highest point, which is $0 \, \text{m/s}$.
• $u$ is the initial velocity, $19.0 \, \text{m/s}$.
• $a$ is the acceleration due to gravity, $-9.8 \, \text{m/s}^2$.
• $s$ is the displacement from the original position.

Solving for $s$:

$0 = (19.0)^2 + 2(-9.8) s$

$0 = 361 - 19.6 s$

$19.6 s = 361$

$s = \frac{361}{19.6} = 18.42 \, \text{m}$

Since the ball is thrown from a height of $5.0 \, \text{m}$, the maximum height from the ground is:

$\text{Maximum height} = 5.0 + 18.42 = 23.42 \, \text{m}$

But since the question asks for the rise from its original position, the answer is $s = 18.42 \, \text{m}$.

(b) Time to Reach Highest Point

To find the time to reach the highest point, we use the first equation of motion:

$v = u + at$

Where:

• $v = 0 \, \text{m/s}$ (velocity at the highest point),
• $u = 19.0 \, \text{m/s}$,
• $a = -9.8 \, \text{m/s}^2$.

Solving for $t$:

$0 = 19.0 - 9.8 t$

$9.8 t = 19.0$

$t = \frac{19.0}{9.8} = 1.94 \, \text{s}$

(c) Time to Hit the Ground

Now we calculate the time taken for the ball to hit the ground from its highest point. At the highest point, the ball has a vertical displacement of $23.42 \, \text{m}$ from the ground.

Using the equation:

$s = ut + \frac{1}{2} a t^2$

Here:

• $s = -23.42 \, \text{m}$ (displacement downwards, hence negative),
• $u = 0 \, \text{m/s}$ (initial velocity at the highest point),
• $a = -9.8 \, \text{m/s}^2$.

$-23.42 = 0 + \frac{1}{2} (-9.8) t^2$

$-23.42 = -4.9 t^2$

$t^2 = \frac{23.42}{4.9}$

$t = \sqrt{\frac{23.42}{4.9}} \approx 2.18 \, \text{s}$

(d) Velocity Upon Return to Original Level

The velocity of the ball when it returns to its original level (5 m height) can be found using the equation:

$v^2 = u^2 + 2 a s$

Where:

• $u = 19.0 \, \text{m/s}$ (initial speed),
• $a = -9.8 \, \text{m/s}^2$,
• $s = 0 \, \text{m}$ (since it's back to the original level).

$v = -19.0 \, \text{m/s}$

The velocity is the same magnitude as the initial speed but in the opposite direction (downward).

Final Answers

1. (a) The ball rises $18.42 \, \text{m}$ from its original position.
2. (b) The time to reach the highest point is $1.94 \, \text{s}$.
3. (c) The time to hit the ground after reaching the highest point is $2.18 \, \text{s}$.
4. (d) The velocity when it returns to the original level is $-19.0 \, \text{m/s}$.

Would you like any details or clarifications on this solution?

Related Questions:

1. How would the maximum height change if the initial speed were doubled?
2. What would be the time of flight if the ball were thrown from the ground level?
3. How would air resistance affect the maximum height reached by the ball?
4. What is the total displacement of the ball when it returns to its initial level?
5. How much time in total does the ball spend in the air before hitting the ground?

Tip: In free-fall problems, always carefully consider the direction and sign of acceleration, as it affects all calculations.