Get Answers to all your Questions

header-bg qa

A plane is inclined at an angle \alpha =30^{0} with respect to the horizontal. A particle is projected with a speed u=2ms^{-1}from the base of the plane , making an angle\theta =15^{0} 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:\left ( Take \, \, g=10ms^{-2} \right )
Option: 1 20cm
Option: 2 18cm
Option: 3 26cm
Option: 4 14cm

Answers (1)

best_answer

As we know on inclined plane the range is given by

R=\frac{2u^{2}}{g}\frac{\sin \beta \cos \left ( \beta +\alpha \right )}{\cos ^{2}\alpha }

\beta =15^{0}\, \, ;\alpha =30^{0}\, \, ;u=2m/s

R=\frac{2\times 2^{2}\sin 15^{0}\cos \left ( 45 \right )}{10\times \cos \left ( 30^{0} \right )}

     = 20 cm \, \, approx

Posted by

Ritika Jonwal

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE