JEE Mains · Physics · STD 11 - 3.2 motion in plane
A plane is inclined at an angle \(\alpha = 30^o\) with respect to the horizontal. A particle is projected with a speed \(u = 2\,ms^{-1},\) from the base of the plant, making an angle \(\theta = 15^o\) with respect to the plane as shown in the figure. The distance from the base at which the particle hits the plane is close to ........ \(cm\) (Take \(g = 10\,ms^2\) )
- A \(18\)
- B \(14\)
- C \(26\)
- D \(20\)
Answer & Solution
Correct Answer
(D) \(20\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} T = \frac{{2\,u\,\sin \theta }}{{g\,\cos \alpha }}\\ R\, = u\,\cos \theta \,T - \frac{1}{2}\,g\,\sin \,\alpha \,{T^2}\\ \,\,\,\,\, = \,\frac{{u\,\cos \,\theta \,2u\,\sin \,\theta }}{{g\,\cos \,\alpha }} - \frac{{g\,\sin \,\alpha }}{2}\frac{{4{u^2}{{\sin…
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