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 time T when both balls are at the same height, we can use the kinematic equation for uniformly accelerated motion:

$s = ut + \frac{1}{2}at^2$

For Ball A:

• Initial velocity, $u_A = 2 \text{ m/s}$
• Initial height, $s_A(0) = 50 ext{ m}$
• Acceleration due to gravity, $a = -9.8 \text{ m/s}^2$

The height of Ball A after time T is given by:

$s_A(T) = 50 + 2T - \frac{1}{2}(9.8)T^2 = 50 + 2T - 4.9T^2$

For Ball B:

• Initial velocity, $u_B = 20 \text{ m/s}$

The height of Ball B after time T is:

$s_B(T) = 20T - \frac{1}{2}(9.8)T^2 = 20T - 4.9T^2$

Setting these two heights equal, we have:

$50 + 2T - 4.9T^2 = 20T - 4.9T^2$

This simplifies to:

$50 + 2T = 20T$

Rearranging gives:

$50 = 20T - 2T$ $50 = 18T$ $T = \frac{50}{18} \approx 2.777...$

Thus, rounding to two significant figures, we find: $T \approx 2.8 \text{ seconds}$