Question

12) A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. What was the ball’s initial velocity?

Answer 1

Answer on Question #58684, Physics / Mechanics | Relativity |

12) A ball is thrown straight up. It passes a 2.00-m-high window $7.50\,\mathrm{m}$ off the ground on its path up and takes $1.30\,\mathrm{s}$ to go past the window. What was the ball's initial velocity?

Solution:

An object in free fall experiences an acceleration $g$ of $-9.81\,\mathrm{m/s^2}$. (The - sign indicates a downward acceleration.)

Given:

$$a = g = -9.81\,\mathrm{m/s^2} \text{ is acceleration}$$ $$h = 2.00\,\mathrm{m}$$ $$y = 7.50\,\mathrm{m}$$ $$t = 1.30\,\mathrm{s}$$

$v_0 = ?$ is initial speed

Suppose $v_1$ is the velocity at 7.50 m off the ground, and $v_2$ is the velocity at 9.50 m off the ground.

Hence,

$$v_2 - v_1 = gt$$ $$v_2^2 - v_1^2 = 2gh$$

From first equation,

$$v_2 = v_1 + gt$$

Substituting to second equation

$$(v_1 + gt)^2 - v_1^2 = 2gh$$ $$v_1^2 + 2v_1gt + g^2t^2 - v_1^2 = 2gh$$ $$2v_1gt = 2gh - g^2t^2$$ $$v_1 = \frac{h}{t} - \frac{gt^2}{2} = \frac{2.00}{1.30} + \frac{9.81 \cdot 1.30^2}{2} = 9.83\,\mathrm{m/s}$$

The initial velocity is

$$v_0^2 = v_1^2 - 2gy = 9.83^2 - 2 \cdot (-9.81) \cdot 7.50 = 243.78$$ $$v_0 = \sqrt{243.78} = 15.61\,\mathrm{m/s}$$

Answer: $15.61\,\mathrm{m/s}$

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