Hulk Strong

The data given in the question is approximate, so we can only find an approximate value for the force. Also, looking at the options, we see that they differ by a factor of 100. We only require an estimate of the final answer and we have a wide window for our final answer. We can afford to make some approximations along the way to make our calculations simpler.

The Hulk jumps up to a height of $h = 120$ meters. Using conservation of energy for the Hulk just after he loses contact with the ground and at the highest point of the jump, we get $\dfrac{1}{2} mv^2 = mgh$ . This gives us the expression for Hulk's initial velocity: $v = \sqrt{2gh} = \sqrt{2400}$ m/s. We see that $\sqrt{2400}$ is close to $\sqrt{2500}$ , which is approximately $50$ m/s.

The Hulk's momentum was initially zero, since he was at rest. The Hulk's mass is about $m = \SI{500}{\kilo \gram}$ , so the Hulk's momentum just after he lost contact from ground was about $mv = 2.5 \times 10^4$ kg m/s. This change in momentum took place in about $\SI{0.3}{\second}$ , so the average force exerted by the Hulk on the ground was

$\begin{aligned} \text{Average Force } &= \dfrac{\text{change in momentum}}{\text{time taken}} \\ & = \dfrac{\Delta p}{\Delta t} \\ & = \dfrac{2.5\times 10^4}{0.3} \text{ Newtons} \\ & = \dfrac{2.5\times 10^5}{3} \text{ Newtons} \end{aligned}$

We see that $2.5$ is approximately $3$ , therefore $\dfrac{2.5}{3} \approx 1$ . The average force exerted on ground by the Hulk is approximately $1 \times 10^5$ Newtons.

Note: The approximations that we have made are valid since they are small compared to the freedom we have (a factor of 100). If we had taken the Hulk's mass as 100 kg, or 10000 kg, we would still have gotten the correct answer.

Simply force = mass*acceleration

We know mass = 500

By using 2nd equation s=ut+1/2at^2 (s = 120,u=0,t=0.3)

By substituting the values

120=0+1/2 a 0.09 240 100/9 80 100/3 8 10^3/3 Therefore 8 10^3/3 is the acceleration

Force = 500 8 10^3/3 =4*10^5/3 Approximately it is equal to 1 *10^5 N

We can solve the problem if we make the right assumptions that no work input is wasted as heat or sound due to drag friction. We then have that P.E gained = K.E lost This gives us the permission to say \frac {1}{2}mv^2 = mgh Where v is the speed at which he must lunch himself. To find v, we use Newton's second law. \frac {∆ P }{∆t} = F The change in momentum is given by ∆P j = mv j i.e in the upward direction. But v = \ sqrt {2gh} from energy conversion, so F =- \sqrt {2gh}\frac {m}{∆t} j Seeing that ∆t is 0.3s, and h = 120m and m = 500kg, also \sqrt{6} is approximately 2.4 We can arrive at F = 80,000N Sorry, my latex code isn't functioning well, if this can be edited, I'd be thankful.