From the edge of a cliff, a 0.55 kg projectile is launched with an initial kinetic energy of 1550 J. The projectile's maximum upward displacement from the launch point is 131 m.
a, What is the horizontal component of its velocity?
b, What was the vertical component of its velocity just after launch
c, At one instant during its flight the vertical component of its velocity is 65 m/s. At that time, what is its vertical displacement from the launch point? (Indicate the direction with the sign of your answer.)
So far:
I have vert = 1550J = 1/2mv^2 + mmg131meters = 55.4
horiz = 1550J = 1/2mvi^2
= 75.1
Not sure what next.
Let's call $v_x$ and $v_y$ the two components of the projectile's velocity. If the vertical velocity is $0$ after the projectile reaches height $h$, the initial vertical velocity is given by $\sqrt{2gh}$.
When the projectile starts, we know that $\frac{1}{2}m(v_x^2+v_y^2) = 1550$ J. Since you know $v_y$, you can compute $v_x$, which never changes because there are no horizontal forces.
For the last part, note that the given velocity of $65$ m/s is higher in absolute value than the initial $v_y$. (If the opposite had been true, the projectile would have achieved that speed twice: once going up and once coming down.)
Apply the simple formulae for speed and displacement in constantly accelerated motion to find the answer.