Answer to Question #342399 in Math for Dicorlysky

Question #342399

Find the initial speed which a projectile must be subjected to give it a maximum horizontal range of 490m. Assume the acceleration due to gravity as g=10m/s



1
Expert's answer
2022-05-19T18:45:31-0400

"x(t)=v_{0x}t=v_0\\cos \\alpha t"

"y(t)=v_{0y}t-\\dfrac{gt^2}{2}=v_0\\sin \\alpha t-\\dfrac{gt^2}{2}"

"t_{fligt}=\\dfrac{2v_{0y}}{g}=\\dfrac{2v_{0}\\sin \\alpha}{g}"

Then


"x_{flight}=v_0\\cos \\alpha(\\dfrac{2v_{0}\\sin \\alpha}{g})=\\dfrac{v_{0}^2\\sin(2 \\alpha)}{g}"

A maximum horizontal range will be at "\\alpha =45\\degree." We have


"v_0=\\sqrt{x_{max}g}"

"v_0=\\sqrt{490m(10m\/s^2)}=70m\/s"


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