Question II

II) A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and its highest point is 140 m above the launch point. (This is a projectile motion.) (a) What was the vertical component of its velocity just right after launch? (10 pts). What is the horizontal component of its velocity? (10 pts)

Solution:
From conservation of energy, the energy at the initial launch equals the energy at the highest point of the trajectory.
KE_i = KE_f + PE_f
$1550 J = \frac{1}{2} m v_{x}^{2} + mg(140m)$
since there is no $\hat{y}$ component of the velocity when the projectile reaches its highest point. There is still an $\hat{x}$ component then however. Solving for v_x, we find
.275 v_x² = 1550 J - 754.6 J
or v_x = 53.8 m/s
Now, looking at the initial kinetic energy,
$1550 J = \frac{1}{2}mv_{i}^{2}$
or v_i = 75.1 m/s
From Pythagorean's theorem,
$v_{i} = \sqrt{v_{x}^{2} + v_{y}^{2}}$
$75.1 = \sqrt{(53.8)^{2} + v_{y}^{2}}$
Solving for v_y, we find
v_y = 52.4 m/s