A Tough Projectile

Classical Mechanics Level 3

A ball is projected with a speed of 20 m/s 20 \text{ m/s} at an angle of 3 0 ∘ 30^\circ from a point on the top of a very high tower. The time after which its velocity becomes perpendicular to the velocity of projection (take g = 10 m / s 2 m/s^2 ) is

If you have the solution then please post it with an explanation.

2 sec 4 sec never 0.5 sec

Sahasra Ranjan by Sahasra Ranjan , 20, India

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1 solution

Dipesh Shivrame
Jun 13, 2016

u = 10\sqrt{3} i + 10 j let the time after which the velocity becomes perpendicular to its initial velocity be t. v= 10\sqrt{3} i + (100-100t)

Finally u.v =0 solving we get t= 4 sec

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Its a very basic problem of kinematics. It requires very basics of projectile and vector knowledge.

Priyanshu Mishra - 4 years, 10 months ago

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