5. A ball is thrown with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball when it gets to 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? (Careful!) (g) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?
Question
- A ball is thrown with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball when it gets to 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? (Careful!) (g) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?
Solution
(a) The velocity of the ball at its highest point is 0 m/s. This is because at the highest point, the ball stops moving upwards and is about to start falling downwards.
(b) The velocity of the ball 1 second before it reaches its highest point depends on the initial speed and the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s². So, if we denote the initial speed as v0, then the velocity 1 second before the highest point is v0 - 9.8 m/s.
(c) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is (v0 - 9.8 m/s) - v0 = -9.8 m/s.
(d) The velocity of the ball 1 second after it reaches its highest point is -9.8 m/s. This is because the ball has started to fall and is accelerating downwards due to gravity.
(e) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is -9.8 m/s - 0 = -9.8 m/s.
(f) The change in velocity during the 2-second interval is the final velocity after 2 seconds minus the initial velocity, which is (-9.8 m/s) - (v0 - 9.8 m/s) = -19.6 m/s.
(g) The acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity is -9.8 m/s². This is the acceleration due to gravity, and it is always acting on the ball, regardless of its velocity.
Similar Questions
Every word problem, there should be a conclusion. therefore add a conclusion to the following answer. "(a) The velocity of the ball at its highest point is 0 m/s. This is because at the highest point, the ball stops moving upwards and is about to start falling downwards. (b) The velocity of the ball 1 second before it reaches its highest point depends on the initial speed and the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s². So, if we denote the initial speed as v0, then the velocity 1 second before the highest point is v0 - 9.8 m/s. (c) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is (v0 - 9.8 m/s) - v0 = -9.8 m/s. (d) The velocity of the ball 1 second after it reaches its highest point is -9.8 m/s. This is because the ball has started to fall and is accelerating downwards due to gravity. (e) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is -9.8 m/s - 0 = -9.8 m/s. (f) The change in velocity during the 2-second interval is the final velocity after 2 seconds minus the initial velocity, which is (-9.8 m/s) - (v0 - 9.8 m/s) = -19.6 m/s. (g) The acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity is -9.8 m/s². This is the acceleration due to gravity, and it is always acting on the ball, regardless of its velocity."
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