A small stone is thrown straight up from the ground with an initial speed of 20 m s⁻¹ - Leaving Cert Physics - Question b - 2015

To calculate the height reached by the stone after two seconds, we can use the equation of motion: $s = ut + \frac{1}{2} a t^2$ where:

• $s$ = height,
• $u$ = initial velocity = 20 m/s,
• $a$ = acceleration due to gravity = -9.8 m/s² (negative because it acts downward),
• $t$ = time = 2 s.

Substituting the values:

1. First part: Calculate the displacement due to initial velocity:
   $s_1 = ut = 20 \times 2 = 40 \text{ m}$

2. Second part: Calculate the displacement due to the acceleration:
   $s_2 = \frac{1}{2} (-9.8) (2^2) = \frac{1}{2} (-9.8) (4) = -19.6 \text{ m}$

3. Now sum both displacements to get the total height: $s = s_1 + s_2 = 40 - 19.6 = 20.4 \text{ m}$

Therefore, the height reached after two seconds is 20.4 meters.