Answer to Question #120140 in Mechanics | Relativity for Arif

Question #120140
A stone is thrown from the level ground (as in projectile motion). It is observed that the maximum height of
its flight is equal to its horizontal range. Find the launch angle θ0.
1
Expert's answer
2020-06-04T10:04:13-0400

Let, "H" is the maximum height,"R" is the horizontal range, "u" is initial velocity and "\\theta" is the projectile angle.

Now, we have given "H=R" , we have to find "\\theta" .

Since, we know that

"H=\\frac{u^2\\sin^2(\\theta)}{2g}"

And

"R=\\frac{u^2\\sin(2\\theta)}{g}"

Thus,

"H=R\\implies \\sin^2(\\theta)=2\\sin(2\\theta)\\\\\n\\implies \\sin^2(\\theta)=4\\sin(\\theta)\\cos(\\theta)\\\\\n\\implies \\tan(\\theta)=4\\implies\\theta=\\tan^{-1}(4)\\approx 76^{\\circ}"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS