Question

A ball rolls off the top of the roof of a building that is 13 meters tall. Calculate the amount of time it takes for it to hit the ground.

Answer 1

Answer:

The ball takes $t=1 \cdot 917sec$ to hit the ground.

Explanation:

• The second equation of motion represents the total distances travelled by an object in a time interval of $\Delta t$ with an initial speed of $u$ and acceleration $a$.

• To find the time, ball takes to hit the ground, use the formula: $s = ut + \frac{1}{2}a{t^2}$

Where, $t$ is time, $u$ is initial velocity, $a$ is acceleration and$s$ is displacement.

• In this case, $a = g = 9 \cdot 8m/se{c^2}$.

• Placing the value of the given initial velocity, $u=0cm/s$ and displacement,$s = 13m$ in the above formula.

$& \therefore s = ut + \frac{1}{2}g{t^2} \\& \Rightarrow 13 = 0 \cdot t + \frac{1}{2} \times 9 \cdot 8 \times {t^2}$

$& \Rightarrow 13 = 4 \cdot 9{t^2} \\& \Rightarrow {t^2} = \frac{{13}}{{4.9}} \\& \Rightarrow t = 1 \cdot 917sec \\\end{align}\]$

• Hence, ball takes $t=1 \cdot 917sec$ to hit the ground.

Learn more about second equation of motion here:

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