Question

an eagle is flying due east at 8.9 m/s carrying gopher in itstalons. The gopher manages to break free at a height of 12m. Whatis the magnitude of the gophers velocityas it reaches the ground?note: effects of air resistance are not included in thiscalculation

Answer 1

x-component is constant in a projectile. So what's important now isthe y-component:
Since the eagle is moving to the east only, meaning it only has anhorizontal velocity, the initial vertical velocity is zero.
Δy = vot + (1/2)at2
=(1/2)at2...since the initial velocity is zero
12m =(1/2)(-9.8m/s²)t²
t = √(2.448979592)
=1.564921593s

Now use this to find the final vertical velocity:
v = v_o + at
=0 + (-9.8)(1.564921593)
=-15.33623161m/s

Now, combine this to the horizontal velocity:
v = √(x² + y²)
=√((8.9)² +(-15.33623161)²)
=17.73161019m/s

That is the final speed. If you still need the final direction:
θ = tan^-1(y/x)
=tan^-1(-15.33623161 / 8.9)
=-59.87233113

So, the final velocity is:
17.73m/s, -59.87^o