1. 0.5g*t^2 = 2010 m.
4.9t^2 = 2010.
t = 20.3 s. = Fall time.
D = Xo*t. = 193m/s * 20.3s = 3909 m.
2. V=sqrt(Xo^2+Yo^2)=sqrt(193^2+58^2) = 202 m/s.
3. Vo*t + 0.5g*t^2 = 2010 m.
58*t + 4.9*t^2 = 2010.
4.9t^2 + 58t - 2010 = 0.
Use Quadratic Formula.
t = 15.2 s. = Fall time.
D = 193m/s * 15.2s = 2934 m.
A plane is flying horizontally with speed
193 m/s at a height 2010 m above the ground, when a package is dropped from the plane.The acceleration of gravity is 9.8 m/s^2.
1.What is the horizontal distance from the release point to the impact point?
Answer in units of m.
2.A second package is thrown downward from the plane with a vertical speed v1 = 58 m/s.
What is the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground?
Answer in units of m/s.
3.What horizontal distance is traveled by this package?
Answer in units of m
5 answers
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