Question

A ball is horizontally projected with a speed v from the top of a plane inclined at an angle 450 with the

horizontal. How far from the point of projection will the ball strike the plane?​

Answer 1

Answer:

$2 \sqrt{2} \frac{ {v}^{2} }{g}$

Step-by-step explanation:

The question: A ball is projected horizontally with a speed v from the top of a plane inclined at an angle of 45 degrees with the horizontal. How far from the point of projection will the ball strike the plane?

Answer:

Please refer to the attached image for the detailed explanation.

Answer 2

The correct answer is $\frac{u^{2} }{g}$.

Given: Speed = v.

Angle = 45°.

To Find: Range of projectile.

Solution:

Range = $\frac{2u^{2}cosqsinq }{g}$    (here q is angle of projection)

= $\frac{2u^{2}\frac{1}{\sqrt{2}}*\frac{1}{\sqrt{2} } }{g}$

= $\frac{u^{2} }{g}$

Hence, the range of projectile is  $\frac{u^{2} }{g}$.

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