Answer to Question #164318 in Mechanics | Relativity for Eugene Agbo

Question

A particle P is let fall vertically from a height to the ground . Another particle Q is projected vertically down 3s later with a velocity of 3m/s from the same height . If both particle reached the ground simultaneously , calculate 

(a)time by each particle to reach the ground

(d) distance travelled by each particle to the ground

Answer 1

$\text{The formula for the magnitude of the displacement }\newline \text{of the body thrown vertically down}:$

$h =V_0t+\frac{gt^2}{2}$

$\text{for particle P:}$

$V_0=0;h=\frac{gt^2}{2}$

$h =\frac{gt^2}{2}$

$\text{In 3 s, particle A:}$

$h_1 = 4.5g\text{ - distance covered in 3 }$

$V_1 =3g -\text{speed A through 3s}$

$\text{then:}$

$h = h_1+V_1t_f+\frac{gt^2_f}{2}\text{ where }$

$t_f =t-3;t- \text{total fall time of particle A}$

$h = 4.5g+3gt_f+\frac{gt^2_f}{2}$

$\text{for particle Q:}$

$V_0=3;h=3t_f+\frac{gt_f^2}{2}$

$\text{equate expression h for A and Q}$

$4.5g+3gt_f+\frac{gt^2_f}{2}= 3t_f+\frac{gt_f^2}{2}$

$t_f=\frac{4.5g}{3-3g}<0$

$t_f<0 \text{this is not a valid value}$

$\text{therefore the problem has no solution}$

Answer: For the given initial conditions, the problem has no solutions