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Q. 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 $\left[g = 10 \, ms^{- 2}\right]$
Question

NTA AbhyasNTA Abhyas 2022

Solution:

Initial velocities of the particle
$\text{u}_{\text{x}}=10cos 53=6 \, m/s$
$\text{u}_{\text{y}}=10sin 53=8 \, m/s$
$6t=1+t+x.\frac{3}{5}$ [where x is the length of the hypotenuse of the wedge]
$5t=1+\frac{3 x}{5}$ (in $\text{x}$ - direction)
$8t-5t^{2}=4.\frac{x}{5}$ (in $\text{y}$ - direction)
$x=\frac{35}{9}m$
Height $=\frac{35}{9}\times \frac{4}{5}=\frac{28}{9}m$