Question

An object is dropped from the height of 19.6 m. Find the distance travelled by object in last second of its journey.

Answer 1

Answer:

14.7 m

Explanation:

In the question,

Height from which the object is dropped, s = 19.6 m

Acceleration due to gravity, g = 9.8 $\frac{m}{s^{2} }$

Also,

We know that, $s=ut+\frac{1}{2}at^{2}$

where, t is the time taken in seconds.

So,

$19.6=(0)(t)+\frac{1}{2}(9.8)(t^{2} )\\ 19.6=4.9t^{2} \\t^{2}=4\\t=2$

(because, u = Initial speed = 0)

Therefore, time taken to reach the ground, t = 2 seconds.

Now,

Also, distance travelled in nth second is given by,

$s_{n} =u+\frac{a}{2}(2n - 1)$

where, n is the second at which we need to calculate the distance travelled.

So,

at, n = 2

$s_{2} =(0)+\frac{9.8}{2}(2(2)-1)=4.9\times3=14.7m$

Therefore, the distance travelled in last second is 14.7 m.