Question

A stone thrown down with a speed $u$ takes a time $t_1$ to reach the ground, while another stone, thrown upwards from the same point with the same speed, takes time $t_2$. The maximum height of the second stone reaches from the ground is :

• A. $1/2160;g{t}_1{t}_2$
• B. $g8(t_1+t_2{)}^2$
• C. $g8(t_1-t_2{)}^2$
• D. $1/2g{t}_2^2$

Answer 1

The correct option is B $g8(t_1+t_2{)}^2$

Case $\left(1\right)$

By conservation of energy the final $K.E$ in both cases will be same as initial height potential energy and initial velocity $(K.E)$ in same in both case. $\Rightarrow V_{f\left(i\right)}=V_{f\left(ii\right)}=V_{f\left(say\right)}$

Equation of motion First: $v=u-g{t}_2$

$\Rightarrow 0=u-g{t}_2|v=0\left(a\right)$ hight point

$\Rightarrow t_2=\frac{u}{g}.........\left(ii\right)$

Fot $t_2:-v_f=0g{t}_2\Rightarrow t_2=\frac{v}{g}....\left(iii\right)$

$\left(ii\right)+\left(iii\right):t_2+t_2=t_2\frac{u+v_f}{g}=\frac{u+u+g{t}_1}{g}$ | from $\left(i\right)$

$\Rightarrow u=\frac{g}{2}(t_2-t_1)......\left(iv\right)$

By conservation of energy:

$m.g.H=\frac{1}{2}.w{v}_f$| where $H-$max height $v_f-$ Final velocity

$\Rightarrow H=\frac{1}{2g}(u+g{t}_1{)}^2$ from $\left(i\right)$

$=\frac{1}{2g}{\left(\frac{g}{2}\right(t_2-t_1)+g{t}_1)}^2$ | from $\left(iv\right)$

$\Rightarrow H=\frac{g}{2}(t_1+t_2{)}^2$

Option $\left(B\right)$