Question

A ball is dropped from a height. If it takes 0.2 sec to cross the last 6 m before hitting the ground, find the height from which it is dropped. Take $g=10 m/s^2$

Answer 1

Hint: When a body is dropped from a particular height, it will travel in a straight line from the point through which it is dropped and will fall vertically downwards. This motion is taking place on the surface of earth hence the only force acting on it will be gravitational force or it’s weight. Hence it is an example of accelerated motion where acceleration is $g=10 m/s^2$.

Formula used:
$s= ut+\dfrac12 a t^2$

Complete step-by-step answer:
Here we are asked about the distance, giving information about the time. Hence we can use any equation of motion to calculate the parameters but the easiest would be using a second equation of motion i.e. $s= ut+\dfrac12 a t^2$.
As it is dropped, the initial velocity of the body is zero.
Hence applying$s= ut+\dfrac12 a t^2$, we get:
$h = 0\times t +\dfrac 12 (-10) t^2 = -5t^2$ . . . . ①
Now it is given to us that it travels 6 m before hitting the ground for 0.2 sec. But here, the initial velocity of the body is not zero for the event, so let it be ‘x’.
Hence using $s= ut+\dfrac12 a t^2$
$-6= x\times 0.2+\dfrac12 (-10) (0.2)^2 = x(0.2) - 0.2$
Or $x= -29 m/s$
Now, this is the velocity at the start of the event. Also the velocity of particle at the start of event is given by equation:
$v= u+at$
$x = 0 + (-10) t$
Hence equating both the equations:
$-10 t = -29$
Hence total time of falling = $\dfrac{-29}{-10} = 2.9 sec$
Now putting value of ‘t’ in equation ①, we get:
$h = -5\times (2.9)^2 = -42.05 m$
Where a negative sign shows that the height is measured in a negative direction as that of the initial point.
Hence the height is 42.05 m.

Note: Students must not get confused by negative and positive signs. We take starting points as origin and then measure quantities with respect to this point. If the direction of measurement of a quantity is towards down or left, then we take negative signs otherwise positive.