At time $t = 0$, two balls A and B are projected vertically upwards - Edexcel - A-Level Maths Mechanics - Question 4 - 2013 - Paper 1

To find the value of $T$, we will use the kinematic equation: $s = ut + \frac{1}{2}at^2$

For ball A, which is projected upwards:

• Initial velocity, $u_A = 2$ m/s
• Initial height, $h_A = 50$ m (above ground)
• Acceleration, $a = -9.8$ m/s$^2$ (due to gravity)

The height of ball A at time $T$ is given by: $s_A = 50 + 2T - \frac{1}{2}(9.8)T^2$

For ball B, starting from the ground:

• Initial velocity, $u_B = 20$ m/s
• Initial height, $h_B = 0$ m

The height of ball B at time $T$ is: $s_B = 20T - \frac{1}{2}(9.8)T^2$

Setting the heights equal at $t = T$ gives: $50 + 2T - 4.9T^2 = 20T - 4.9T^2$

This simplifies to: $50 + 2T = 20T$

Rearranging yields: $50 = 18T$

Thus, we find: $T = \frac{50}{18} \approx 2.777... \approx 2.8 \, \text{s}$