Question

A particle is projected from the ground and simultaneously a wedge starts moving towards the right, as shown in the figure. The maximum height of the wedge for which the particle will not hit the wedge is $\left[g=10m{s}^{-2}\right]$

• A. $\frac{28}{9}m$
• B. $\frac{28}{3}m$
• C. $\frac{7}{3}m$
• D. $\frac{7}{4}m$

Answer 1

Answer: (A)

Initial velocities of the particle
${\text{u}}_{\text{x}}=10cos53=6m/s$
${\text{u}}_{\text{y}}=10sin53=8m/s$
$6t=1+t+x.\frac{3}{5}$ [where x is the length of the hypotenuse of the wedge]
$5t=1+\frac{3x}{5}$ (in $\text{x}$ - direction)
$8t-5t^2=4.\frac{x}{5}$ (in $\text{y}$ - direction)
$x=\frac{35}{9}m$
Height $=\frac{35}{9}\times \frac{4}{5}=\frac{28}{9}m$