Understanding Vertical Motion and Time of Flight

This problem involves analyzing the motion of a ball thrown vertically upwards under the influence of gravity. When an object is thrown upwards, it decelerates due to gravity until it reaches its highest point, where its instantaneous velocity is zero. It then accelerates downwards, returning to the starting point.

The total time the ball spends in the air, from the moment it's thrown until it returns to the ground, is called the time of flight. For vertical motion starting and ending at the same level, the time taken to reach the highest point is equal to the time taken to fall back from the highest point to the initial level.

Calculating Time to Reach Maximum Height

Given that the total time of flight is 10 seconds, the time taken for the ball to reach its maximum height is half of the total time of flight.

Time to reach maximum height (\(t_{\text{up}}\)) = \(\frac{\text{Total time of flight}}{2}\)

\(t_{\text{up}} = \frac{10 \text{ s}}{2} = 5 \text{ s}\)

At the maximum height, the final velocity of the ball (\(v\)) is 0 m/s.

Applying Kinematic Equations to Find Initial Velocity

We can use the first equation of motion under constant acceleration to find the initial velocity. The equation is:

\(v = u + at\)

Where:

• \(v\) is the final velocity.
• \(u\) is the initial velocity (what we need to find).
• \(a\) is the acceleration. In this case, it's the acceleration due to gravity (\(g\)). Since the ball is moving upwards, gravity acts downwards, opposing the motion, so the acceleration is negative (\(a = -g\)).
• \(t\) is the time taken. Here, we use the time taken to reach the maximum height.

Given:

• \(v = 0 \text{ m/s}\) (velocity at maximum height)
• \(a = -g = -10 \text{ m/s}^2\)
• \(t = 5 \text{ s}\) (time to reach maximum height)
• \(u = ?\)

Substitute the values into the equation:

\(0 = u + (-10 \text{ m/s}^2) \times (5 \text{ s})\)

\(0 = u - 50 \text{ m/s}\)

Now, solve for \(u\):

\(u = 50 \text{ m/s}\)

So, the velocity with which the ball was thrown up is 50 m/s.

Summary of Steps

1. Identify the total time of flight.
2. Calculate the time taken to reach the maximum height (half of the total time).
3. Recognize that the velocity at the maximum height is zero.
4. Use the kinematic equation \(v = u + at\).
5. Substitute the known values (\(v=0\), \(a=-g\), time to max height) and solve for \(u\).

Revision Table: Vertical Motion Concepts

Additional Information: Kinematic Equations for Constant Acceleration

The following equations are fundamental for solving problems involving constant acceleration, like vertical motion under gravity:

• \(v = u + at\) (Relates final velocity, initial velocity, acceleration, and time)
• \(s = ut + \frac{1}{2}at^2\) (Relates displacement, initial velocity, acceleration, and time)
• \(v^2 = u^2 + 2as\) (Relates final velocity, initial velocity, acceleration, and displacement)
• \(s = \frac{(u+v)}{2}t\) (Relates displacement, initial and final velocities, and time)

Where:

• \(s\) is the displacement.
• \(u\) is the initial velocity.
• \(v\) is the final velocity.
• \(a\) is the constant acceleration.
• \(t\) is the time interval.

For vertical motion, 'a' is replaced by 'g' (or -g depending on the direction) and 's' is often replaced by 'h' (height) or 'y' (vertical position).