We can use the kinematic equation:
y = yi + vi*t + (1/2)*a*t^2
where
y = final height (relative to the point of release)
yi = initial height (100m above the point of release)
vi = initial velocity (0 m/s since the package is dropped from rest)
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken to reach the ground
Since the package is dropped, we can assume that its initial velocity is 0 m/s. We can rearrange the equation to solve for t:
t = sqrt((2*yi)/a)
t = sqrt((2*100)/9.8) = 4.04 s
Now we can use another kinematic equation to find the horizontal distance travelled by the package:
x = xi + vi*t
where
x = horizontal distance travelled
xi = initial horizontal position (unknown)
vi = horizontal velocity (40 m/s)
t = time taken (4.04 s)
x = vi*t = (40 m/s)*4.04 s = 161.6 m
Therefore, the package hits the ground 161.6 m horizontally away from the point of release.
A rescue plane drops a package of emergency rations to a stranded explorer. The plane is travelling at 40m/s and at a height of 100m above the ground. Where does the package hit the ground relative to the point it was released?
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1 answer