Question

The speed of a projectile at its maximum height is Root 3/2 times its initial speed. If the range of the projectile is p times the maximum height attained by it , then p=

Answer 1

using some concept related to 2-d motion we can solve easily...

see attachment.....

◆remember always velocity at highest point is ucos@

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HOPE IT WILL HELP U

Answer 2

Given that,

The speed of a projectile at its maximum height is $\dfrac{\sqrt 3}{2}$ times its initial speed.

To find,

The value of p if range of the projectile is p times the maximum height attained by it.

Solution,

Initial speed of the projectile is $u\cos\theta$.

According to question,

$u\cos\theta=\dfrac{\sqrt3}{2}u\\\\\cos\theta=\dfrac{\sqrt3}{2}\\\\\theta=30^{\circ}$

Range of projectile is :

$R=\dfrac{u^2\sin2\theta}{g}\\\\R=\dfrac{u^2\sin2(30)}{g}\\\\R=\dfrac{\sqrt3 u^2}{2g}$

Maximum height of projectile is :

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{u^2\sin^2(30)}{2g}\\\\H=\dfrac{u^2}{8g}$

According to question,

$\text{Range}=p\times \text{H}\\\\\dfrac{\sqrt3 u^2}{2g}=p\times \dfrac{u^2}{8g}\\\\p=4\sqrt{3}$

So, the value of p is $4\sqrt{3}$.