Question

An arrow is fired with initial velocity v0 at an angle θ from the top of battlements, a height h above the ground. (Assume θ is measured above the horizontal.)
(a) In terms of h, v0, θ, and g, what is the time at which the arrow reaches its maximum height?
t =


(b) In terms of h, v0, θ, and g, what is the maximum height above the ground reached by the arrow?
hmax =

Answer 1

at the top, vertical velocity is zero.
vf=0=vo*sinTheta -gt
(vo*sinTheta-gt)=0
at the top, t=vo*sinTheta/g

At the top, height hmax
hmax=vo*sinTheta*t-4.8t^2
put the expression for t above into that to find hmax