Question

A ball is projected with a speed of 20m/s at an angle of 30^(@) 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=10m/s^(2) ) is (A) 0.5sec (B) 2sec (C) 4sec (D) never PLS SOMEONE GIVE THE RIGHT ANSWER

Answer 1

Sorry dude I don't know

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Answer 2

In this question ,

initial velocity of projection(u) = 20 m/s

$\alpha = 30$

We know that through out the motion the velocity along the X direction remains the same

$u \cos( \alpha ) = v \sin( \alpha )$

$v = u \tan( \alpha )$

$v = 20 \sqrt{3}$

Now using the second equation of motion in y direction

$- vy = uy - gt$

$- v \: cos \alpha = u \: sin \alpha - gt$

$- 20 \sqrt{3} \cos(30) = 20 \sin(30) - 9.8t$

$- 20 \sqrt{3} \times \frac{ \sqrt{3} }{2} = \frac{20}{2} - 9.8t$

$- 30 - 10 = - 9.8t$

$- 4 0 = - 9.8t$

$t = \frac{40}{9.8}$

$t = 4 \: sec$