Question

A ball is thrown vertically upwards. It was observed at a height $h$, twice with a time interval $\Delta t$ . The initial velocity of the ball is

• A. $\sqrt{8gh+g^2(\Delta t{)}^2}$
• B. $\sqrt{8gh+{\left(\frac{g\Delta t}{2}\right)}^2}$
• C. $\frac{1}{2}\sqrt{8gh+g^2(\Delta t{)}^2}$
• D. $\sqrt{8gh+4g^2(\Delta t{)}^2}$

Answer 1

Answer: (C)

$\frac{1}{2}\sqrt{8gh+g^2(\Delta t{)}^2}$

If the ball is reaching height h then, displacement$=h$

Time taken$=t_1$... (say)

Acceleration$=-g$

Initial velocity$=u_1$

$\therefore$from the equations of motion

$S=ut+\frac{1}{2}a{t}^2\therefore h=u_1{t}_1+\frac{1}{2}(-g)t_1^2.....\left(i\right)$

If the ball is reaching height $2h$then, displacement$=2h$

Time taken$=t_1+△t$

Acceleration$=-g$

Initial velocity$=u_2$

Therefore from equations of motion

$2h=u_2(t_1+△t)-\frac{1}{2}g{(t_1+△t)}^2.....\left(ii\right)$

Also we know from the equations of motion that

$v=u+at\therefore 0=u_1-g{t}_1....\left(iii\right)$

& $0=u_2-g(t_1+△t).....\left(iv\right)$

$\therefore u_1=g{t}_1,u_2=g{t}_1+g△t=u_1+g△t$

$\therefore$From equations $\left(i\right),\left(ii\right),\left(iii\right),\left(iv\right)$

we get$\left(ii\right)-\left(i\right)=h=\{(u_1+g△t)(t_1+△t)-\frac{1}{2}g{(t_1+△t)}^2-u_1{t}_1-\frac{1}{2}g{t}_1^2\}$

$\therefore u_1=\frac{1}{2}\sqrt{8gh+g^2{\left(△t\right)}^2}$ Thus option C