Standing on the ground from 3.0 m from a building, you want to throw a package from your 1.5- shoulder level to someone in a window 4.2 m above the ground. At what speed and angle should you throw the package so it just barely clears the window sill?
So first I set up my parameters:
initial x=0
initial y=1.5
x velocity=?
y velocity= 0
x acceleration=0
y acceleration=-9.8m/s^2
final x=3.0 m
final y= 2.7 m (4.2-1.5)
final x velocity=?
final y velocity=0
t=.93s (I found this by solving the equation y=1/2at^2+vt)
are my parameters correct, if not where did I go wrong?
Note that your vertical displacement is $d_y = 4.2 - 1.5 = 2.7$. Now there are infinitely many parabolic trajectories that will solve the problem. To fix this, let's interpret "so it just barely clears the window sill" to mean "so that the final vertical velocity is $v_{fy} = 0$", as you correctly figured out. Then since $a_y = -9.8$ and we want to solve for the initial vertical velocity $v_{iy}$, we use the following kinematics formula: $$ v_{fy}^2 = v_{iy}^2 + 2a_yd_y \implies v_{iy} = \sqrt{v_{fy}^2 - 2a_yd_y} = \sqrt{0^2 - 2(-9.8)(2.7)} = 7.2746... $$ We can also solve for time: $$ v_{fy} = v_{iy} + a_yt \iff t = \frac{v_{fy} - v_{iy}}{a} = \frac{0 - 7.2746...}{-9.8} = 0.7423... $$ So we can now handle the horizontal components: $$ v_{ix} = \frac{d_x}{t} = \frac{3.0}{0.7423...} = 4.0415... $$ Finally, our initial velocity is: $$ v_i = \sqrt{v_{ix}^2 + v_{iy}^2} = \sqrt{(4.0415...)^2 + (7.2746...)^2} \approx 8.32 $$ and our angle is: $$ \theta = \tan^{-1}(v_{iy}/v_{ix}) = \tan^{-1}(7.2746.../4.0415...) \approx 61^\circ $$