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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.

user416503
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    whats mmg131meters? By the way, start with variables $v$ and $\theta$ and use kinematic equations and conservation of energy. – jonsno Mar 28 '17 at 05:08
  • mg X 131 meters... Which kinematic? – user416503 Mar 28 '17 at 05:18
  • You are "double-counting" the kinetic energy. First, you assume that all kinetic energy comes from vertical velocity. Then you assume that it all comes from horizontal velocity. In reality, the initial kinetic energy comes from the absolute value of the velocity. – Fabio Somenzi Mar 28 '17 at 05:32

1 Answers1

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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.

Fabio Somenzi
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