A ball is dropped from the edge of a roof. It takes $0.1 s$ to cross a window of height $2.0 m$ . Find the height of the roof above the top of the window.

9.8 m

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19.6 m

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29.4 m

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39.2 m

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Solution

The correct option is B $19.6 m$

Let $A B$ be the window, and suppose the roof is at a height $x$ above $A$ .
Also, suppose it takes a time $t_{1}$ for the ball to reach A.
The velocity of the ball at A is
$v_{1} = 0 + g t_{1} = (9.8 {m/s}^{2}) t_{1}$
Now, consider the motion of the ball from $A$ to $B$ .
Hence, the initial velocity is $v_{1}$ the distance covered is $2 m$ , and the time taken is $0.1 s$ . We have
$s = u t + \frac{1}{2} a t^{2}$
$2.0 m = v_{1} (0.1 s) + \frac{1}{2} \times (9.8 {m/s}^{2}) \times (0.1 s)^{2} = (9.8) t_{1} (0.1) + 0.049$
$2.0 - 0.049 = (0.98) t_{1}$
$t_{2} = \frac{1.951}{0.98} \approx 2 s .$
The height $x$ is
$x = \frac{1}{2} g t_{1}^{2} = \frac{1}{2} \times (9.8) \times (2)^{2} = 19.6 m$
The roof is at a height $19.6 m$ above the top of the window.

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A ball is dropped from the edge of a roof. It takes 0.1 s to cross a window of height 2.0 m. Find the height of the roof above the top of the window.

A ball is dropped from the edge of a roof. It takes $0.1 s$ to cross a window of height $2.0 m$ . Find the height of the roof above the top of the window.