Question

A body projected obliquely with velocity 19.6m/s has kinetic energy at the maximum height equal to 3 times its potential energy there. Its position after 1 second of projection from the ground is(h=maximum height)

Answer 1

Explanation:

Initial velocity u= 19.6 m/s

after every 1 sec of projectile angle made by  particle with horizontal is given by

t= u sinθ/g

1= 19.6 x sinθ/9.8

9.8/19.6= sin θ

sinθ=1/2

θ= 30°

Therefore position of particle after 1 sec is given by

Hmax= U² sin² θ/2g

         =19.6x 19.6 sin² 30/2x 9.8

         = 19.6 x 1/4

         =4.9 m

Answer 2

Answer:

Explanation:

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