A spring-loaded gun can fire a projectile to a height h if it is fired straight up. if the same gun is pointed at an angle of 45◦ from the vertical, what maximum height h45◦ can now be reached by the projectile

Answers

When firing straight up: v^2 = u^2 - 2gh, where v = final velocity = 0, u = initial velocity, g = gravitational acceleration, h = maximum height attained. Then, 0 = u^2 - 2gh u = Sqrt (2gh) ---- (1) When firing at 45°, Initial velocity, U = u Sin 45 = Sqrt (2gh)·Sin 45 Maximum height, H = U^2*(Sin Ф)^2/2g substituting; H = [Sqrt (2gh)·Sin 45]^2*(Sin 45)^2]/2g H = [2gh*(Sin 45)^2*(Sin 45)^2]/2g H = [h*(Sin 45)^4] = h/4 Therefore, maximum height when the gun fires at 45° is a quarter of maximum height when the gun fires vertically up.


0 0
Only authorized users can leave an answer!
Can't find the answer?

If you are not satisfied with the answer or you can’t find one, then try to use the search above or find similar answers below.

Find similar answers


More questions