Question

A particle is projected from a horizontal plane with a velocity of 8 root 2 m/s at an angle. At the highest point it’s velocity is found to be 8 m/s. It’s range will be?

Answer 1

Answer:

Hey mate your answer is here

That is 12.8

Answer 2

The range of the projectile is 12.8 meters.

Explanation:

Given that,

Initial velocity of projection,$u=8\sqrt{2}\ m/s$

Velocity at the highest point, v = 8 m/s

We need to find the range of the projectile.

Using trigonometric equation,

$v=u\ cos\theta$

$8=8\sqrt{2} \ cos\theta$

$\theta=45^{\circ}$

Now using the formula of the range of projectile we get :

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

$R=\dfrac{(8\sqrt{2})^2 sin(90)}{10}$

R = 12.8 meters

So, the range of the projectile is 12.8 meters. Hence, this is the required solution.

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