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 with which the ball strikes the ground, we again use the equation of motion:

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

In this case:

• Initial velocity, $u = 14.7 \, \text{m/s}$
• Displacement from the highest point to the ground, which is $s = -49 \text{ m}$ (downwards)
• Acceleration due to gravity, $a = 9.8 \, \text{m/s}^2$

Substituting these values:

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

Calculating gives:

$v^2 = 216.09 - 960.4 = -744.31$

Since we are interested in the absolute value of velocity:

Calculating $v$ gives:

$v \approx 34.3 \, \text{m/s}$

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

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

Where:

• $s = -49 \, \text{m}$
• Initial velocity, $u = 14.7 \, \text{m/s}$
• Acceleration, $a = -9.8 \, \text{m/s}^2$

Setting up the equation:

$-49 = 14.7t - \frac{1}{2}(9.8)t^2$

This simplifies to the quadratic equation:

$4.9t^2 - 14.7t - 49 = 0$

Using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 4.9$, $b = -14.7$, and $c = -49$ gives:

Calculating: $t = \frac{14.7 \pm \sqrt{(-14.7)^2 - 4(4.9)(-49)}}{2(4.9)}$

This results in: $t = 5 \, \text{s}$

Thus, the total time from projection to when it strikes the ground is 5 seconds.