Explain the step by step process to solve the question on trajectory motion. A ball is thrown vertically upward and reaches maximum height of 20ms⁻¹ find The speed of projection when it is at a height of 10 meters taking g to be 10ms⁻²

We use the information that the ball reaches a maximum height of $20 \text{ m}$. At its maximum height, the ball's instantaneous velocity is $0 \text{ m/s}$.
We use the kinematic equation that relates initial velocity ($u$), final velocity ($v$), acceleration ($a$), and displacement ($s$):

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

Given:

• Final velocity at maximum height, $v = 0 \text{ m/s}$
• Displacement (maximum height), $s = 20 \text{ m}$
• Acceleration due to gravity, $a = -g = -10 \text{ m/s}^2$ (negative because gravity acts downwards, opposing the upward motion)

Substitute these values into the equation:

$$(0 \text{ m/s})^2 = u^2 + 2(-10 \text{ m/s}^2)(20 \text{ m})$$ $$0 = u^2 - 400 \text{ m}^2/\text{s}^2$$ $$u^2 = 400 \text{ m}^2/\text{s}^2$$ $$u = \sqrt{400 \text{ m}^2/\text{s}^2}$$ $$u = 20 \text{ m/s}$$

The initial speed of projection is $20 \text{ m/s}$.

Now we use the initial speed ($u$) we just calculated to find the speed ($v$) of the ball when it is at a height of $10 \text{ m}$.
We use the same kinematic equation:

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

Given:

• Initial speed, $u = 20 \text{ m/s}$ (from Step 1)
• Displacement (height), $s = 10 \text{ m}$
• Acceleration due to gravity, $a = -g = -10 \text{ m/s}^2$

Substitute these values into the equation:

$$v^2 = (20 \text{ m/s})^2 + 2(-10 \text{ m/s}^2)(10 \text{ m})$$ $$v^2 = 400 \text{ m}^2/\text{s}^2 - 200 \text{ m}^2/\text{s}^2$$ $$v^2 = 200 \text{ m}^2/\text{s}^2$$ $$v = \sqrt{200 \text{ m}^2/\text{s}^2}$$ $$v = \sqrt{100 \times 2} \text{ m/s}$$ $$v = 10\sqrt{2} \text{ m/s}$$

The speed of the ball when it is at a height of 10 meters is $\boxed{10\sqrt{2} \text{ m/s}}$.