Answer:
a)The package strikes 256.2 m in the ground relative to the point directly below where it was released
b) The horizontal component will not change it remains same as 41 m/s
c) Vertical component of velocity = 61.41 m/s
Explanation:
a) Consider the vertical motion of plane,
We have equation of motion, s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Displacement, s = 192 m
Acceleration, a = 9.81 m/s²
Substituting
s = ut + 0.5 at²
192 = 0 x t + 0.5 x 9.81 x t²
t = 6.26 seconds
Now we need to find horizontal distance traveled in 6.26 seconds by the package.
We have equation of motion, s = ut + 0.5 at²
Initial velocity, u = 41 m/s
Time, t = 6.26 s
Acceleration, a = 0 m/s²
Substituting
s = ut + 0.5 at²
s = 41 x 6.26 + 0.5 x 0 x 6.26²
s = 256.52 m
The package strikes 256.2 m in the ground relative to the point directly below where it was released
b) The horizontal component will not change it remains same as 41 m/s
c) We have equation of motion, v = u+ at
Initial velocity, u = 0 m/s
Time, t = 6.26 s
Acceleration, a = 9.81 m/s²
Substituting
v = u+ at
v = 0 + 9.81 x 6.26 = 61.41 m/s
Vertical component of velocity = 61.41 m/s
