Question

A stuntman jumps from the roof of a building to the safety net below. How far has the stuntman fallen after 2 seconds?

Answer 1

Hint :When a body is falling towards the ground or rising up to a height, in both cases acceleration due to gravity will pull it downward with a constant magnitude of $ {\text{9}}{\text{.8 m}}{{\text{s}}^{ - 2}} $ . As the acceleration is constant, all the kinematics equations will be valid here.

Complete Step By Step Answer:
Firstly, in order to understand which kinematic equation will yield us the result for the above problem, writing down the given data below,
As he jumps from the building this means that his initial velocity was zero. We will write it as,
Initial velocity, $ u = 0{\text{ m}}{{\text{s}}^{ - 1}} $
And as given, Time, $ t = 2\operatorname{s} $
Now, we have to find the distance he covered in the given time,
Distance travelled, $ s = ? $
As per the data, we now know that the second kinematic equation should be used which is as follows,
 $ s = ut + \dfrac{1}{2}a{t^2} $
Here, s is the distance travelled.
u is the initial velocity
a is acceleration
and, t is time taken.
As acceleration here is the acceleration due to gravity which acts downward so by convention we will use $ {\text{a = - 9}}{\text{.8 m}}{{\text{s}}^{ - 2}} $ , where minus sign denotes the downward direction.
Hence, using the data above, our equation will become,
 $ s = (0 \times 2) + \dfrac{1}{2} \times ( - 9.8) \times {2^2} $
 $ s = \dfrac{1}{2} \times ( - 9.8) \times {2^2} $
 $ s = - 19.6{\text{ m}} $
So, the distance travelled by the stuntman after $ 2 $ seconds will be $ - {\text{ }}19.6{\text{ m}} $ , and the minus sign here indicates the downward direction.

Note :
If we haven’t used the negative sign for acceleration in the first place then distance travelled would have been in positive, which would have denoted the distance in positive y direction which is impossible as one cannot go up after jumping off the building. So, while solving these kinds of problems, one should pay keen attention to the sign convention.