A body is projected vertically upward. Show that the

(i) distance covered by it in the first ‘n’ seconds of ascent is equal to the distance covered by it in the first ‘n’ seconds of its descent.

(ii) the distance covered by it in the last ‘n’ seconds of ascent is equal to the distance covered by it in the first ‘n’ seconds of its descent.

Question

A body is projected vertically upward. Show that the

(i) distance covered by it in the first ‘n’ seconds of ascent is equal to the distance covered by it in the first ‘n’ seconds of its descent.

(ii) the distance covered by it in the last ‘n’ seconds of ascent is equal to the distance covered by it in the first ‘n’ seconds of its descent.

Accepted Answer

(i) Find the velocity v1 of a body projected upwards after [latex] t=\left(t_{1}-n\right)[/latex] seconds [latex] t_{1}[/latex] = time of ascent = [latex] \left(u_{1} / g\right)[/latex] using equation [latex] v_{1}=u_{1}-g t[/latex]

(ii) [latex] v_{1}[/latex] is the initial velocity in the beginning of last n seconds

(iii) Find [latex] s_{1} [/latex] by substituting [latex] v = 0[/latex] and [latex] u = v_{1}[/latex] in equation [latex] v^{2}=u^{2}-2 g s _{1}[/latex]

(iv) Find the displacement [latex] s_{2}[/latex] for the body falling from rest for n seconds using equation [latex] s =u t+1 / 2 g t^{2}[/latex]

(v) Compare [latex] s_{1}[/latex] and [latex] s_{2}[/latex] .

Question : A body is projected vertically upward. Show that the (i) dis...