A body is projected obliquely with velocity 19.6m/s has its 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)?
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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
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
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