A ball is projected vertically upwards with speed 21 m s⁻¹ from a point A, which is 1.5 m above the ground - Edexcel - A-Level Maths Mechanics - Question 5 - 2007 - Paper 1

To find the speed of the ball when it reaches the ground, we can use the same kinematic equation:

$v^2 = u^2 + 2as$

We take the reference point at the ground (0 m) and calculate the total distance fallen from point A (1.5 m + 22.5 m = 24 m) with initial velocity as 0:

$v^2 = 0 + 2 \times 9.8 \times 24$

Calculating v gives:

$v = \sqrt{470.4} \approx 22 ext{ ms}^{-1}$

Thus, the speed of the ball as it reaches the ground is approximately 22 m s⁻¹.

To find the total time of flight, we can use the equation:

$v = u + at$

Using the final velocity obtained (22 m s⁻¹) and substituting:

$22 = 21 - 9.8t$

Rearranging gives:

$t = \frac{21 - 22}{-9.8} = \frac{1}{9.8} \approx 0.102 ext{ s}$

Next, to find the time to rise to the maximum height,

Using the equation:

$v = u + at$

We find:

$0 = 21 - 9.8t_{up}$

Thus:

$t_{up} = \frac{21}{9.8} \approx 2.14 ext{ s}$

Now, the total time of flight includes both the ascent and descent, hence:

$t_{total} = t_{up} + t_{down} \\ 0.102 + 2.14 = 2.24 ext{ s}$

Finally, the time between when the ball is projected from A and when it reaches the ground is approximately 4.4 seconds.