Question

5. A ball is projected to attain the maximum range.
If the height attained is H, the range is​

Answer 1

Answer:

range=height×cos theta

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Answer 2

4H

Explanation:

The range formula in projectile motion is given as,

 $R=\frac{u^2sin2\theta}{g}$

Here, u is the initial velocity and $\theta$ is the angle of projection.

The height formula in projectile motion,

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

For maximum range, $\theta=45^0$, so

$R_{max}=\frac{u^2sin2\times45^0}{g}$

$R_{max}=\frac{u^2}{g}$                 ( because, $sin 90^0=1$)

Maximum height,

$H=\frac{u^2(sin45^0)^2}{g}$

$H=\frac{u^2}{4g}$                      ( because, $sin 45^0=\frac{1}{\sqrt{2} }$)

Now comparing maximum range and maximum height, we get

$R_{max}=4H$

Thus, the maximum range will be 4H.

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