A project is thrown at an angle of `theta = 45^circ` to the horizontal , reaches a maximum height of 16 m , then Options: A . &nbspits velocity at the highest point is zero B . &nbspits range is 64 m C . &nbspit is thrown at an angle of `30^circ` , its range will decrease D . &nbsp(B) and (C) both the correct

A project is thrown at

Question -

A project is thrown at an angle of `theta = 45^circ` to the horizontal , reaches a maximum height of 16 m , then

Options:

A . its velocity at the highest point is zero

B . its range is 64 m

C . it is thrown at an angle of `30^circ` , its range will decrease

D . (B) and (C) both the correct

Answer: Option D

The maximum height is

`H = (u^2 sin^2 theta)/(2g)`

or `16 = (u^2 sin^2 45^circ)/(19.6)`

`:.` `u = sqrt(16 xx 2 xx 19.6)`

But `R = (u^2 sin 2 theta)/(g)`

= `(16 xx 2 xx 19.6 sin 90^circ)/(9.8)` = 64 m.

when `theta = 45^circ, sin 2 theta` = 1

But when `theta = 30^circ, sin 2 theta = sqrt(3)/2` = 0.866

so, at `theta = 30^circ, ` range is lesser than that of at `theta = 45^circ`

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