# A projectile is launched over a horizontal surface in such a manner that its maximum height is 1/4 of its horizontal range. Determine the la

Question

A projectile is launched over a horizontal surface in such a manner that its maximum height is 1/4 of its horizontal range. Determine the launch angle.

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1 year 2021-09-05T07:11:28+00:00 1 Answers 12 views 0

The launch angle $$\theta=45^o$$.

Explanation:

We know,

Horizontal distance travelled in projectile motion , $$R=\dfrac{u^2sin2\theta}{g}$$.

Also, Maximum height ,$$H_{max}=\dfrac{u^2sin^2\theta}{2g}$$.

Now, according to question maximum height is 1/4 of its horizontal range.

$$H_{max}=\dfrac{1}{4}\times R$$

$$\dfrac{u^2 sin^2\theta}{2g}=\dfrac{1}{4}\times \dfrac{u^2 sin\ 2\theta}{g}$$

Simplifying above expression.

We get , $$\theta=45^o$$

Hence, this is the required solution.