Let's break down the problem step-by-step to find the unknown, $T$, where it seems to be referring to the time at which the two balls are at the same vertical height again.

Step 1: Define the Motion of Both Balls

We have two balls:

• Ball A is projected from a height of 50 m with an initial velocity of 2 m/s upwards.
• Ball B is projected from the ground with an initial velocity of 20 m/s upwards.

We will use the standard kinematic equation for vertical motion under gravity:

$h(t) = h_0 + v_0 t - \frac{1}{2} g t^2$ Where:

• $h(t)$ is the height at time $t$.
• $h_0$ is the initial height.
• $v_0$ is the initial velocity.
• $g$ is the acceleration due to gravity, which is approximately $9.8 \, \text{m/s}^2$.

Step 2: Write the Equations of Motion for Both Balls

Ball A:

• Initial height $h_0 = 50 \, \text{m}$
• Initial velocity $v_0 = 2 \, \text{m/s}$

$h_A(t) = 50 + 2t - \frac{1}{2} (9.8)t^2$

Ball B:

• Initial height $h_0 = 0 \, \text{m}$
• Initial velocity $v_0 = 20 \, \text{m/s}$

$h_B(t) = 0 + 20t - \frac{1}{2} (9.8)t^2$

Step 3: Use the Given Information at $t = 7 \, \text{seconds}$

At $t = 7$ seconds, both balls are at the same height. So:

$h_A(7) = h_B(7)$

Substitute $t = 7$ into both equations:

$h_A(7) = 50 + 2(7) - \frac{1}{2}(9.8)(7^2)$

$h_B(7) = 20(7) - \frac{1}{2}(9.8)(7^2)$

Simplify the equations:

For Ball A:

$h_A(7) = 50 + 14 - \frac{1}{2}(9.8)(49) = 50 + 14 - 240.1 = -176.1 \, \text{m}$

For Ball B:

$h_B(7) = 140 - \frac{1}{2}(9.8)(49) = 140 - 240.1 = -100.1 \, \text{m}$

Step 4: Solve for $T$

From these, the first observations show their mistake