36. A ball is tossed with enough speed straight up so that it is in the air several seconds.
a) What is the velocity of the ball when it reaches its highest point?
b) What is its velocity 1 s before it reaches its highest point?
c) What is the change in its velocity during this 1 s interval?
d) What is its velocity 1 s after it reaches its highest point?
e) What is the change in velocity during this 1 s interval?
f) What is the change in velocity during the 2 s interval?
g) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?
To answer these questions, we need to understand the basic principles of projectile motion and the equations that govern it. Let's go step by step:
a) The velocity of the ball at its highest point is zero. This is because at the highest point, the ball momentarily comes to a stop before starting to fall back down due to the acceleration of gravity.
b) To find the velocity of the ball 1 s before reaching its highest point, we need to take into account the acceleration due to gravity. The ball is moving upwards at this point, so the velocity will be decreasing. We can use the equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. In this case, the final velocity is what we need to find, the initial velocity is the speed at which the ball was thrown upwards, the acceleration is the acceleration due to gravity (which is approximately -9.8 m/s^2), and the time is 1 s. Substituting these values into the equation, we can find the velocity.
c) The change in velocity during this 1 s interval can be found by subtracting the initial velocity from the final velocity. So, the change in velocity is vf - vi.
d) The velocity of the ball 1 s after it reaches its highest point can be found in a similar way as part b. However, now the ball is moving downwards, so the acceleration due to gravity works in the same direction as the motion. The equation can be written as:
vf = vi + at
where the initial velocity would be 0 (since the ball momentarily comes to a stop at its highest point), the acceleration is still -9.8 m/s^2, and the time is 1 s. Solving this equation will give us the final velocity.
e) The change in velocity during this 1 s interval can be found by subtracting the initial velocity from the final velocity, similar to part c.
f) The change in velocity during a 2 s interval can be found by repeating the steps described in parts b and c, but this time using a time interval of 2 s instead of 1 s.
g) The acceleration of the ball during any of these time intervals can be considered constant and equal to the acceleration due to gravity, which is approximately -9.8 m/s^2. At the moment the ball has zero velocity (highest point and 1 s after reaching it), the acceleration is still the same.
By following these steps and using the appropriate equations, you can find the answers to the questions regarding the velocity, change in velocity, and acceleration of the ball during different time intervals.