Question

The ceiling of a hall is 30 m high.A ball is thrown with 60m/s at an angle theeta, so that maximum horizontal distance may be covered. Find the sine of angle of projection

Answer 1

Answer:

Since the height of the room is 30m , it is obvious that the max height has to be 30m.

The velocity of projection is given as 60m/s.

Remember the formula for max height in a projectile and solve for sin(theta).

Please check the answer in the assignment book, hope it is right and you understand.

Answer 2

The sine of angle of projection is $\dfrac{1}{\sqrt{6} }$.

Explanation:

Given that,

The height of the hall, h = 30 m

Initial speed of the ball, u = 60 m/s

The ball is thrown with this speed so that maximum horizontal distance may be covered i.e. it will cover a height of 30 m.

The maximum height of a projectile is given by :

$h=\dfrac{u^2\ \sin^2\theta}{2g}\\\\30=\dfrac{(60)^2\ \sin^2\theta}{2\times 10}\\\\\sin\theta=\sqrt{\dfrac{30\cdot20}{60^{2}}}\\\\\sin\theta=\dfrac{1}{\sqrt{6} }$

So, the sine of angle of projection is $\dfrac{1}{\sqrt{6} }$.

Learn more,

Projectile motion

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