SOLUTION: When an object Is released and falls freely, it's height, h metres, above the ground after t seconds is represented by the relation h= - 0.5 t^2 g + d ,Where G is the accelerat
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Question 1151156: When an object Is released and falls freely, it's height, h metres, above the ground after t seconds is represented by the relation h= - 0.5 t^2 g + d ,Where G is the acceleration due to gravity and D is the initial height of the object before it's released. since the string of gravity varies from planet to planet the values of G is specific to each planet
suppose a rock is dropped from a height of 100 m on Mars and also on Venus. on Mars G= 3.7m/s and Venus, g= 8.9m/s. how long does it take the rock to fall to a height of 25 M on each planet show all your work.
Found 2 solutions by Alan3354, jim_thompson5910:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
suppose a rock is dropped from a height of 100 m on Mars and also on Venus. on Mars G= 3.7m/s and Venus, g= 8.9m/s. how long does it take the rock to fall to a height of 25 M on each planet show all your work.
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h(t) = -gt^2/2 + d
d = 100
Mars g = 3.7m/sec^2
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h(t) = -3.7t^2/2 + 100 = 25
3.7t^2 = 150
Solve for t, ignore the negative value
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Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
For the planet Mars, g = 3.7 m/s^2 is the acceleration of gravity.
d = 100 is the starting height in meters.
h = 25 is the ending height we want the object to be at (ie the object travels 100-25 = 75 vertical meters through the air)
We will plug those values into the equation and solve for t
h = -0.5*t^2*g + d
25 = -0.5*t^2*3.7 + 100
25 = -0.5*3.7*t^2 + 100
25 = -1.85t^2 + 100
25 + 1.85t^2 = 100
1.85t^2 = 100-25
1.85t^2 = 75
t^2 = 75/1.85
t^2 = 40.5405405405405
t = sqrt(40.5405405405405)
t = 6.36714539967013
t = 6.4
When starting at a height of 100 m, it takes approximately 6.4 seconds for the object to fall to a height of 25 meters. In other words, it takes roughly 6.4 seconds for the object to fall 75 meters when starting at a height of 100 meters. This only applies on the planet Mars.
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For Venus, repeat the same basic steps as in the previous part above. The only difference is that g = 8.9 (bigger planet means stronger gravitational pull).
The other values d = 100 and h = 25 remain the same.
h = -0.5*t^2*g + d
25 = -0.5*t^2*8.9 + 100
25 = -0.5*8.9*t^2 + 100
25 = -4.45t^2 + 100
25 + 4.45t^2 = 100
4.45t^2 = 100-25
4.45t^2 = 75
t^2 = 75/4.45
t^2 = 16.8539325842697
t = sqrt(16.8539325842697)
t = 4.1053541362798
t = 4.1
When starting at a height of 100 m, it takes approximately 4.1 seconds for the object to fall to a height of 25 meters on the planet Venus.
For each planet, we are ignoring air resistance because that greatly complicates the problem.
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