Stone A is projected vertically upwards at a speed of 12 m s⁻¹ from a height h above the ground - NSC Physical Sciences - Question 3 - 2017 - Paper 1

Since stone A reaches its maximum height at $12$ m/s upwards, the speed of stone B when stone A reaches maximum height is:

$3v = 12 + 9.8 t$

Substituting in the time calculated in part 3.1:

$3v = 12 + 9.8 \times 1.22$

Calculating gives:

$3v = 12 + 11.96 \approx 23.96$

Thus, we can find v:

$v = \frac{23.96}{3} \approx 7.99 \, \text{m/s}$

Using the equations of motion for stone B. Since stone B is thrown downwards with an initial speed $v$:

$h = v t + \frac{1}{2} a t^2$

Where:

• $v = 7.99 \, \text{m/s}$
• $a = 9.8 \, \text{m/s}^2$
• $t = 1.22 \, \text{s}$

Substituting gives:

$h = 7.99 \times 1.22 + \frac{1}{2} \times 9.8 \times (1.22)^2$

Calculating: $h \approx 9.73 + 7.29 \approx 17.02 \, \text{m}$

To sketch the velocity-time graphs for stones A and B:

• Stone A:

  • Starts with an initial velocity of 12 m/s upwards.
  • Reaches maximum height in approximately 1.22 seconds where velocity is 0 m/s, then falls back down.
  • The graph will show a linear decrease to 0 and then a linear increase downwards.

• Stone B:

  • Starts at the same height, thrown downwards at speed 7.99 m/s.
  • The velocity will increase as it falls, accelerating at 9.8 m/s².
  • The graph will show a linear upward trend starting from 7.99 m/s and increasing as it approaches the ground.