Answer to Question #121373 in Classical Mechanics for Quanda Simplice Nzuning

Question

At a high jump competition, Ashu clears a height of 2.0m. He takes 1.26 s to do the jump (i.e. from the moment he lifts off the ground to when he touches down again).
(i) Sketch a graph showing a variation of the height h/m of the lowest part of his body from the ground with time t/s. Indicate clearly on the graph a few important values of h and corresponding values of t.
(ii) Explain briefly how Ashu’s speed at any height could be determined from the graph.
(iii) Describe the energy changes that take place during the jump.
(iv) In what way is Ashu’s momentum at take-off different from that at touch down. Explain without any calculations.

Answer 1

Remember how height changes with time:

$$h(t)=v_0t-\frac{gt^2}{2}$$

We know that the time Ashu moved up is the same as the time Ashu moved down, 0.63 s. Therefore, substituting this value for t and 2 m for h, we can find the initial velocity:

$$2=v_0\cdot0.63-\frac{9.8\cdot0.63^2}{2}\rightarrow v_0=6.3\text{ m/s}.$$

(i) Sketch the graph:

(ii) The speed at any point can be determined if we draw a tangent to any point and divide the height of the tangent (vertical size along y-axis) by its base (horizontal length along x-axis).

(iii) Initial energy: Maximum kinetic, zero potential. Between 0 and maxium height to the left of the blue vertical kinetic energy decreases while potential increases.

At the highest point: maximum potential, zero kinetic. After the highest point kinetic increases as potential decreases to zero.

(iv) The direction and magnitude of Ashu's velocity vector: at the start it points upward and decreases while after the highest point before Ashu landed the velocity vector points downward and increases in length.