For an object projected from the ground with speed u, horizontal range is two times the maximum height attained by it. The horizontal range of the object is:

Here, we will use the relation between the horizontal range R, acceleration due to gravity g, initial speed u and the maximum height H, which is given as:

R = H ⇒u2sin2θ/g = 2× u2sin2θ/g

⇒ 2sinθcosθ = sin2θ

⇒ tanθ = 2

From this we get the following values:

sinθ =2√5

cosθ = 1√5

Now, by substituting these values to find the horizontal range R:

R = u2×2 sinθ cos θg

⇒ R = u2 × 2 × 2/√×1/√5g

∴R = 4u2/5g

Therefore, we get the required horizontal range R of the object, projected from the ground with speed u.

Was this answer helpful?


0 (0)


Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *




App Now