Question

A 5Kg object is tossed with an initial velocity of 100m/s at an unknown angle, if the object travels to a max height of 450.8m and max horizontal distance of 328.3m, what angle is the object thrown at? What is its maximum gravitational potential energy? Where does it reach it's maximum KE, of 25,000?

Answer 1

well we know max horizontal distance=\[\left( ? \right) u^{2}\sin ^{2\Theta}\left( ? \right)\div \left( 2g \right)\] so substituting values we get angle=53.34

Answer 2

Where did you get the equation you used and are both the question marks the same thing?

Answer 3

well when you throe something due to gravity it travels unusual path (parabolic path) imagine youself throwing a ball aobserve how it travels!
and the eq. hope you get it