Question

When a ball is thrown upwards, the time, $T$ seconds, during which the ball remains in the air is directly proportional to the square root of the height, $h$ metres, reached. We know $T=4.47sec$ when $h=25m$. If the ball is thrown upwards and remains in the air for $5$ seconds, find the height reached (correct to $2$ decimal places)

Answer 1

The statement " $x$ varies directly as $y$ ," means that when $y$ increases, $x$ increases by the same factor. In other words, $y$ and $x$ always have the same ratio that is:

$x=ky$

where $k$ is the constant of variation.

Here, it is given that the time $T=4.47$ sec is directly proportional to the square root of the height $h=25$ m, therefore,

$T=k\sqrt{h}\Rightarrow 4.47=k\sqrt{25}\Rightarrow 4.47=5k\Rightarrow k=\frac{4.47}{5}=0.894$

Thus, the general equation is $T=0.894\sqrt{h}$.

Since $T=5$ sec, therefore,

$5=0.894\sqrt{h}\Rightarrow {\left(5\right)}^2={\left(0.894\right)}^2\times {\left(\sqrt{h}\right)}^2\Rightarrow 25=0.799h\Rightarrow h=\frac{25}{0.799}=31.27$

Hence, if the ball is thrown upwards and remains in the air for $5$ seconds, then the height of $31.27$ m is reached.