A ball is projected vertically upwards with a speed of 14.7 m s⁻¹ from a point which is 49 m above horizontal ground - Edexcel - A-Level Maths Mechanics - Question 6 - 2010 - Paper 1

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

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

This time:

• $u = 14.7$ m/s (upward),
• $a = -9.8$ m/s² (downward),
• $s = 49$ m (the height from which it falls).

Substituting the values:

$v^2 = (14.7)^2 + 2(-9.8)(-49)$

Calculating gives:

$v^2 = 216.09 + 960.4 = 1176.49$

Thus,

$v = \sqrt{1176.49} \approx 34.3 ext{ m/s}$

Hence, the speed with which the ball first strikes the ground is approximately 34.3 m/s.

To find the total time, we can use the following equation:

$v = u + at$

Where:

• $v$ is the final speed (34.3 m/s),
• $u$ is the initial speed (14.7 m/s),
• $a$ is the acceleration due to gravity (-9.8 m/s²), and
• $t$ is time.

Rearranging the equation gives:

$t = \frac{v - u}{a}$

Substituting the values:

$t = \frac{34.3 - 14.7}{9.8} \approx 2.0 ext{ seconds}$

Additionally, to confirm: Using another kinematic equation:

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

Where:

• $s$ is the total height (49 m), leading to:

$49 = 14.7t - 4.9t^2$

This leads to a quadratic resolution confirming that $t \approx 5 ext{ seconds}$. Therefore, the total time from when the ball is projected to when it first strikes the ground is approximately 5 seconds.