Explanation :
It is given that:
Mass of person, m = 64 kg
We know that, F = m a
Where, a is acceleration due to gravity and \(a =\dfrac{v}{t}\)
(i) Force exerted at t = 0 s :
\(\because F = ma\)
At t = 0 s, the body is at rest so no acceleration. The force will be zero.
(ii) At t = 0.5 s, force of gravity will also acts i.e. f = m (a+g)
From the graph it is clear that at t = 0.5 s, velocity is 15 cm/s =0.15 m/s
So,\(F=64\ kg\times (\dfrac{0.15\ cm/s}{0.5\ s} +9.8\ m/s^2)\)
\(F = 646.4\ N\)
(ii) Force at t = 1.1 s :
At t = 1.1 s, v = 28 cm/s = 0.28 m/s
\(F=64\ kg\times (\dfrac{0.28\ m/s}{1.1\ s}+9.8\ m/s^2)\)
\(F =643.2\ N\)
(iii) Force at t = 1.6 s :
At t = 1.6 s, v = 6 cm/s = 0.06 m/s
\(F=64\ kg\times (\dfrac{-0.06 \ m/s}{1.6\ s}+9.8\ m/s^2)\) (decelerating)
\(F =-627.1\ N\)
Hence, this is the required solution .
