In a competition, participants must attempt to throw a ball vertically upwards past point T, marked on a tall vertical pole - NSC Physical Sciences - Question 3 - 2018 - Paper 1

To find the time taken to reach the highest point, we can use the formula:

$t = \frac{v_f - v_i}{a}$

where:

• $v_f = 0 \, m/s$ (final velocity at the highest point)
• $v_i = 7.5 \, m/s$ (initial velocity)
• $a = -9.81 \, m/s^2$ (acceleration due to gravity, negative because it acts downwards)

Substituting the values:

$t = \frac{0 - 7.5}{-9.81} \approx 0.764 \, s$

To find out if the ball passes point T, we must first calculate the maximum height reached by the ball.

The maximum height can be calculated using:

$h = v_i t - \frac{1}{2} a t^2$

Substituting the known values:

$h = 7.5 imes 0.764 - \frac{1}{2} \times 9.81 \times (0.764)^2$

Calculating this gives:

$h \approx 7.5 imes 0.764 - 0.5 \times 9.81 \times 0.583\approx 5.73 \, m$

The initial height of the throw is 1.6 m, so the total height reached is:

$5.73 + 1.6 = 7.33 \, m$

Since 7.33 m > 3.7 m, the ball will indeed pass point T.

The velocity-time graph during the motion will be a straight line starting from the initial velocity at 7.5 m/s and sloping downwards to zero at the highest point.

• The initial velocity is 7.5 m/s (at t=0).
• The final velocity at the highest point is 0 m/s (at t=0.764 s).
• The time taken to reach the highest point is 0.764 s.

The graph would look like:

|   v  |
| 7.5  | 
|      |______
|      |      
|      |     
|      |   
|      | 
|______|______ 
       0   0.764 t

The slope of the line represents the acceleration due to gravity.