Question

A rock is thrown straight up with an initial velocity of 15.0 m/s. Ignore energy lost to air friction. How high will the rock rise?

Answer 1

`h=v_0t- (1)/(2) gt^2\\h=15t-(1)/(2)(9.8) t^(2)\\ \\ v_f=v_0-gt\;\;\;\Rightarrow t= (v_f-v_0)/(-g) = (0-15)/(-9.8) \approx1.53 \\ \\ \Rightarrow h=15(1.53)-4.9(1.53)^2\;\;\;\Rightarrow h=11.48\,m`