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.

Getting Info...
Cookie Consent
We serve cookies on this site to analyze traffic, remember your preferences, and optimize your experience.
It seems there is something wrong with your internet connection. Please connect to the internet and start browsing again.
AdBlock Detected!
We have detected that you are using adblocking plugin in your browser.
The revenue we earn by the advertisements is used to manage this website, we request you to whitelist our website in your adblocking plugin.
Site is Blocked
Sorry! This site is not available in your country.