Question

A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer 1

Step 1:Given data:

A stone is thrown in a vertically upward direction with a velocity of 5 m/s.

The acceleration of the stone during its motion is 10 m/s² in downward.

Initial velocity u = 5 ${\mathrm{ms}}^{-1}$
Final velocity v = 0
$g=10{\mathrm{ms}}^{-2}$

Step 2: Formula used

We know the third equation of motion,

${\mathrm{v}}^2={\mathrm{u}}^2-2\mathrm{gh}$

(here v = final velocity, u = initial velocity, g is the acceleration due to gravity, h is height negative sign appears because motion is in an upward direction and acceleration is in downward direction)

Step 3:Finding height:

Applying the second equation of motion we get,
$$\begin{gathered} 0=5^2-2\times 10\times \mathrm{h} \\ \mathrm{h}=\frac{25}{2\times 10}=1.25\mathrm{m} \end{gathered}$$

Step 4:Finding the time:

Again we know from the first equation of motion,

$v=u+at$

(here v = final velocity, u = initial velocity, g is acceleration, t is time)

Applying the equation we get,
$$\begin{gathered} \mathrm{a}=-\mathrm{g}=-10{\mathrm{ms}}^{-2} \\ 0=5-10\mathrm{t} \\ \mathrm{t}=\frac{5}{10}=0.5\sec \end{gathered}$$

Hence, the height attained by the stone is 1.25m, and time taken to reach is 0.5sec.