Question

A particle is projected from the ground at t = 0 so that on its way it just clears two vertical walls of equal height on the ground. If the particle passes just grazing top of the wall at time $t_{1}and{t}_2$ then calculate

the height of the wall

• A. h = g $(t_1^2+t_2^2+t_1{t}_2)$
• B. h = g $\frac{t_1{t}_2}{2}$
• C. h = g $(t_1+t_2{)}^2$
• D. h = g $\frac{(t_1+t_2)}{2}$

Answer 1

The correct option is B

h = g $\frac{t_1{t}_2}{2}$

Time of Flight = $t_1+t_2$

Assume initial velocity = u

Angle of projection = $\theta$

Now we know time of flight = $\frac{2Usin\theta }{g}=t_1+t_2$

$\Rightarrow$ u sin $\theta$ = $\frac{(t_1+t_2)g}{2}$ .......... (1)

Applying equation of motion in y-axis for part of motion where it reaches wall 1

$u_y$ = u sin $\theta$

$s_y$ = h

t = $t_1$

a = -g

$\Rightarrow$ h = u sin $\theta t_1$ - $\frac{1}{2}g{t}_1^2$ ..........(ii)

Putting equation (i) in (ii) we get

h = $\frac{(t_1+t_2)g}{2}$$t_1$ - $-\frac{1}{2}g{t}_1^2$

$\Rightarrow$ h = $\frac{g}{2}{t}_1$$[t_1+t_2-t_1]$

$h=\frac{1}{2}g{t}_1{t}_2$