A stone is projected from the top of a tower with velocity $$20$$ ms–$$1$$ making an angle $$300$$ with the horizontal. If the total time of flight is $$5$$ s and g $$=$$ $$10$$ ms–$$2$$,

Height from point of projectionTotal height from ground $$=$$ $$75$$ $$+$$ $$5$$ $$=$$ $$80$$ m

Projection Under uniform Acceleration

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Question

A stone is projected from the top of a tower with velocity 20 ms^–1 making an angle 30⁰ with the horizontal. If the total time of flight is 5 s and g = 10 ms^–2,

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Solution

$\large - h = u\sin \theta t - \frac{1}{2}g{t^2}$
$\large \Rightarrow h = - 20\;\frac{1}{2}(5) + \frac{1}{2}(10){(5)^2}$
$\large h = - 50 + 125 = 75m$
Height from point of projection
$\large = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\$ $\large =\frac{(20)^2(\frac{1}{2})^2}{{(2)}(10)}=\frac{100}{20}=5\;m$
Total height from ground = 75 + 5 = 80 m

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

A stone is projected from the top of a tower with velocity 20 ms–1 making an angle 300 with the horizontal. If the total time of flight is 5 s and g = 10 ms–2,

Height from point of projectionTotal height from ground = 75 + 5 = 80 m

Ready to Test Your Skills?

free study material

A stone is projected from the top of a tower with velocity 20 ms^–1 making an angle 30⁰ with the horizontal. If the total time of flight is 5 s and g = 10 ms^–2,