Question

A particle is projected from ground as shown in figure. Particle and wedge start moving simultaneously.

Maximum height of inclined so that particle will not hit the inclined is : $(g=10m/s^2)$

• A. $\frac{28}{9}m$
• B. $\frac{21}{9}m$
• C. $\frac{28}{12}m$
• D. $\frac{21}{12}m$

Answer 1

The correct option is A $\frac{28}{9}m$

$\overrightarrow{v_p}=6\hat{i}+8\hat{j},\overrightarrow{v_w}=\hat{i}\overrightarrow{v_{p/w}}=5\hat{i}+8\hat{j}$

Let the particle just touches the topmost point then the following two equations of verticle displacement and horizontal displacement would be as

$h=8t-\frac{1}{2}g{t}^2.....\left(1\right)1+\frac{3h}{4}=5t.......\left(2\right)\Rightarrow t=\frac{2}{3}s\Rightarrow h=8\times \frac{2}{3}-\frac{1}{2}g\times {\left(\frac{2}{3}\right)}^2=\frac{28}{9}m$