A boy throws a ball vertically upwards. It rises to a maximum height, where it is momentarily at rest, and then falls back to his hands.

Question 10

A boy throws a ball vertically upwards. It rises to a maximum height, where it is momentarily at rest, and then falls back to his hands.

Which row gives the acceleration of the ball at various stages in its motion? (Take vertically upwards as positive. Ignore air resistance.)

rising                at maximum height                 falling

A          –9.81 m s^-2                        0                                  +9.81 m s^-2

B          –9.81 m s^-2                  –9.81 m s^-2                              –9.81 m s^-2

C         +9.81 m s^-2                 +9.81 m s^-2                             +9.81 m s^-2

D         +9.81 m s^-2                       0                                  –9.81 m s^-2

Reference: Past Exam Paper – November 2011 Paper 11 Q8 & Paper 13 Q9 & November 2015 Paper 12 Q7

Solution:

Answer: B.

The acceleration of free, g is a constant and is always directed towards the surface (downwards).

Since the positive direction is taken as being vertically upwards, the value of g should be negative. So, with the sign, g = – 9.81 m s^-2.

So, the value of g is negative and remains the same at any position.

Since air resistance is ignored, the resultant acceleration on the ball is always g in a downward direction whether the ball is rising, stationary or falling.

As the ball is rising, it moves upwards with a certain speed which decreases with time since the acceleration of free fall is always downwards. Motion is in one direction while the acceleration on the ball is in the opposite direction. So, the ball decelerates.

 
At the maximum height, the speed of the ball is momentarily zero but the force of gravity is still present. So, the acceleration of free fall is still – 9.81 m s^-2. Then, as the ball falls, it accelerates downwards with an acceleration equal to g.