To solve this problem, we need to find the total time from when the ball is thrown upwards to when it first hits the ground. Let's break this down:

Given:

• Initial velocity ($u$) = 14.7 m/s (upwards)
• Initial height ($h_0$) = 19.6 m above the ground
• Acceleration due to gravity ($g$) = 9.8 m/s² (downwards)
• Final height when the ball hits the ground ($h_f$) = 0 m

Key equation:

We can use the kinematic equation that relates displacement, initial velocity, time, and acceleration: $h = h_0 + ut - \frac{1}{2} g t^2$ Where:

• $h$ is the final height (0 m, since it hits the ground),
• $h_0$ is the initial height (19.6 m),
• $u$ is the initial velocity (14.7 m/s upwards),
• $g$ is the acceleration due to gravity (9.8 m/s² downwards),
• $t$ is the time to be solved.

This simplifies to: $0 = 19.6 + 14.7 t - \frac{1}{2} (9.8) t^2$ $0 = 19.6 + 14.7 t - 4.9 t^2$

Now we have a quadratic equation: $4.9 t^2 - 14.7 t - 19.6 = 0$

We can solve this quadratic equation using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where:

• $a = 4.9$,
• $b = -14.7$,
• $c = -19.6$.

Let me calculate the value of $t$.The solutions for time ($t$) are:

• $t_1 = 4.0$ seconds
• $t_2 = -1.0$ seconds

Since time cannot be negative, we discard $t_2 = -1.0$ seconds. Therefore, the total time from when the ball is thrown to when it hits the ground is $t = 4.0$ seconds.

Would you like further details or have any additional questions?

Here are 5 related questions to explore:

1. What is the maximum height the ball reaches before descending?
2. How would the time change if the initial height were different?
3. How would the result differ if the ball was thrown from ground level?
4. What is the velocity of the ball just before it hits the ground?
5. How does the direction of gravity influence projectile motion?

Tip: When solving quadratic equations in physics, always check if the solutions are physically meaningful (e.g., discarding negative times).