wodajewode
Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

If U=53m/s and find the launch angle such that the maximum height of the projectile is equal to three times the horizontal range​

Sagot :

Answer:

t = V0 sin theta / g      time to get max height

h = Voy t - 1/2 g t^2 = Vo^2 sin^2 theta / 2 g          max height

R = V0^2 sin 2 theta / g      range formula

h = 3 R        given

Vo^2 sin^2 theta / 2 g = 3 * V0^2 sin 2 theta / g

sin^2 theta = 6 * sin 2 theta = 12 * sin theta cos theta

tan theta = 12

theta = 85.2 deg

Can check:  

Vx = 8.37

Vy = 99.6     assuming V0 =100

t = 99.6 / 9.8 = 10.2 sec     to reach max height

Sx (total range) = 2 * 10.2 * 8.37 = 170 m

Sy = 1/2 * 9.8 * (10.2)^2 = 510 m

Height = 3 * range

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.