Question

A particle is projected vertically upward with a velocity of 15 m/s from the point O. Find

the:

(i) maximum height reached.

(ii) time taken for it to return to O.

(iii) time taken for it to be 25 m below O.

Answer 1

According to Newton's second law

$$m{dv\over dt}=F$$

By the Law of conservation of energy

$$K+P=const$$

Let $y$ denote the height of the particle above the ground.

Let $v(t)$ denote the velocity of the particle at time $t$

$${mv_0^2\over 2}+0={mv^2\over 2}+mgy$$

$$m{dv\over dt}=-mg$$

$$v(t)=v_0-gt$$

(i) maximum height reached. 

$$v=0, y=y_{max}$$

$${mv_0^2\over 2}+0=0+mgy_{max}$$

$$y_{max}={v_0^2\over 2g}$$

$$y_{max}={(15m/s)^2 \over2(9.8m/s^2)}\approx11.480\ m$$

(ii) Let $t_2=$ time taken for it to return to $0.$  

When the partice reaches maximum height

$$v(t_1)=0=v_0-gt_1$$

$$t_1={v_0\over g}$$

$$t_2=2t_1={2v_0\over g}$$

$$t_2={2(15m/s)\over 9.8m/s^2}\approx3.061\ s$$

(iii) time taken for it to be 25 m below O.

$$v(t)={dy\over dt}$$

$y(0)=0, v(t)=v_0-gt$

$$y=\int v(t)dt=\int(v_0-gt)dt=y(0)+v_0t-{gt^2\over 2}=$$ $$=v_0t-{gt^2\over 2}$$

Given $y=-25\ m$

$$-25=15t_3-{9.8t_3^2\over 2}, t_3\geq0$$

$$4.9t_3^2-15t_3-25=0$$

$$t_3={15\pm\sqrt{(-15)^2-4(4.9)(25)}\over 2(4.9)}=$$

$$={15\pm\sqrt{715}\over 9.8}$$

Since $t_3\geq0,$ we take

$$t_3={15+\sqrt{715}\over 9.8}\approx4.259\ s$$

The particle will be 25 m below approximately $4.259$ seconds after start.