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

Answers ( )

Helga · Answer 1 · 2021-09-05T07:12:36+00:00

Answer:

The launch angle [tex]\theta=45^o[/tex].

Explanation:

We know,

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

Also, Maximum height ,[tex]H_{max}=\dfrac{u^2sin^2\theta}{2g}[/tex].

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

[tex]H_{max}=\dfrac{1}{4}\times R[/tex]

[tex]\dfrac{u^2 sin^2\theta}{2g}=\dfrac{1}{4}\times \dfrac{u^2 sin\ 2\theta}{g}[/tex]

Simplifying above expression.

We get , [tex]\theta=45^o[/tex]

Hence, this is the required solution.