# A ball is thrown vertically upwards from a window that is 3.6m above the ground.The ball's initial speed is 2.8m/s and the acceleration due to gravity is 9.8m/s^2. What is the ball's speed when it hits the ground?

Jul 20, 2015

Speed upon impact: 8.9 m/s

#### Explanation:

You know that your ball is being thrown from a height of 3.6 m with an initial velocity of 2.8 m/s.

At the top of its motion, i.e. at maximum height, the velocity of the ball will be zero. This means that you can determine the height it travels before starting its decent by using the initial velocity

underbrace(v_"top"^2)_(color(blue)("=0")) = v_i^2 - 2 * g * h_"up"

This is equivalent to

${v}_{i}^{2} = 2 \cdot g \cdot {h}_{\text{up" => h_"up}} = {v}_{i}^{2} / \left(2 \cdot g\right)$

${h}_{\text{up" = (2.8^2 "m"^cancel(2) * cancel("s"^(-2)))/(2 * 9.8cancel("m") * cancel("s"^(-2))) = "0.40 m}}$

The ball will be at a total height of

$h = {h}_{i} + {h}_{\text{up" = 3.6 + 0.40 = "4.0 m}}$

Now use the same equation again, only this time take into account the fact that the initial velocity, i.e. the velocity at maximum height, is zero.

v_f^2 = underbrace(v_"top"^2)_(color(blue)("=0")) + 2 * g * h

${v}_{f} = \sqrt{2 \cdot g \cdot h} = \sqrt{2 \cdot 9.8 \text{m" * "s"^(-2) * 4.0 "m}}$

${v}_{f} = \textcolor{g r e e n}{\text{8.9 m/s}}$