A stone is projected vertically upward with a speed of 30ms-1 form the top of a tower of height 50m. neglecting air resistance. determine the maximum height it reached form the ground. (g = 10ms⁻¹)

The stone is projected vertically upward with an initial velocity $u = 30 \, \text{m/s}$. At its maximum height, its final velocity $v = 0 \, \text{m/s}$. The acceleration due to gravity is $g = 10 \, \text{m/s}^2$, acting downwards, so we use $a = -g = -10 \, \text{m/s}^2$ for upward motion.
Using the kinematic equation $v^2 = u^2 + 2as$:

$$0^2 = (30 \, \text{m/s})^2 + 2(-10 \, \text{m/s}^2)s$$ $$0 = 900 \, \text{m}^2/\text{s}^2 - 20 \, \text{m/s}^2 \cdot s$$ $$20s = 900$$ $$s = \frac{900}{20}$$ $$s = 45 \, \text{m}$$

This is the height the stone reaches above the top of the tower.

The tower's height is $50 \, \text{m}$. The maximum height from the ground is the sum of the tower's height and the height reached above the tower.

$$\text{Maximum height from ground} = \text{Height of tower} + s$$ $$\text{Maximum height from ground} = 50 \, \text{m} + 45 \, \text{m}$$ $$\text{Maximum height from ground} = 95 \, \text{m}$$

The maximum height reached from the ground is $\text{95 m}$.