91Ó°ÊÓ

When analyzing the motion of a thrown ball, it's essential to understand how to calculate its velocity. The key equation used for this purpose is a kinematic equation:

\[ v^2 = v_{1}^{2} + 2gh \]

Here, \(v\) is the final velocity just before the ball strikes the ground, \(v_{1}\) is the initial velocity, \(g\) is the acceleration due to gravity (approximately \(9.8 \, m/s^2\)), and \(h\) is the height from which the ball is thrown. This equation allows us to find the speed of the ball right before it hits the ground, regardless of whether it was thrown up or down initially.

To calculate the velocity in part (a) of the exercise, if the ball is thrown straight down with an initial velocity, we substitute in the known values and solve for \(v\):

\[ v = \sqrt{v_{1}^{2} + 2gh} \]

By plugging the initial speed \(v_{1}\) and the height \(h\) into the equation, we can determine the ball's speed at the moment it is about to make contact with the ground.

Determining the time of flight, or the duration for which the ball is in the air, involves solving a quadratic equation derived from kinematic principles. The primary equation used here is:

\[ h = v_{1}t + \frac{1}{2}gt^2 \]

Here, \(t\) represents the time, and the other symbols retain their previously defined meanings. To find the time till the ball hits the ground, we need to solve this quadratic equation for \(t\). This expands to:

\[ \frac{1}{2}gt^2 + v_{1}t - h = 0 \]

Solving it through the quadratic formula, we get:

\[ t = \frac{-v_{1} \pm \sqrt{v_{1}^{2} + 2gh}}{g} \]

We consider only the positive root of this equation, as time cannot be negative. Solving for \(t\) enables us to derive the time the ball takes to hit the ground in part (b) of the exercise.

Gravity is a fundamental force acting on the ball throughout its flight. It constantly accelerates the ball towards the Earth with an acceleration value denoted as \(g\), which equals approximately \(9.8 \, m/s^2\). When calculating the ball's motion, this value is crucial in determining both the velocity and time of flight.

In part (c) and (d) of the exercise, even when the ball is thrown upwards initially, gravity decelerates it until it momentarily stops and then accelerates it downwards again. Due to the symmetric nature of free fall, the final velocity just before hitting the ground remains the same as in part (a):

\[ v = \sqrt{v_{1}^{2} + 2gh} \]

However, the time of flight increases since the ball has to travel upwards before coming back down. This demonstrates the consistent influence of gravitational acceleration throughout the ball’s entire journey.