Question

A boy throws a stone slant wise from a point on the ground with such a speed that it clears the top of a tree and lands on the ground after 4s.if g=10m/s, the height of the tree is

Answer 1

The height of the tree is 20 m

Explanation:

Let the angle of projection be $\theta$ and initial velocity $u$

Given

Time $t=4$ s

If the range is $R$

Then

$R=\frac{u^2\sin2\theta}{g}$

Also

$R=u\cos\theta\times t$

Thus,

$\frac{u^2\sin2\theta}{g}=u\cos\theta\times t$

$\implies u^2\times 2\sin\theta\cos\theta=u\cos\theta\times gt$

$\implies 2u\sin\theta=gt$

$\implies u\sin\theta=\frac{10\times 4}{2}$

$\implies u\sin\theta=20$

If the maximum height covered is $H$

Then height of the tree = $H$

$H=\frac{u^2\sin^2\theta}{2g}$

$\implies H=\frac{20^2}{2\times 10}$

$\implies H=20$ m

Hope this answer is helpful.

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