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pshichka [43]
3 years ago
11

Provides some background for this problem. a ball is thrown vertically upward, which is the positive direction. a little later i

t returns to its point of release. the ball is in the air for a total time of 7.70 s. what is its initial velocity? neglect air resistance.
Physics
1 answer:
julsineya [31]3 years ago
6 0
For an object thrown vertically upward, its time of flight is equal to the equation below:

t = 2v/g
where
t is the time of flight
v is the initial velocity
g is the acceleration due to gravity equal to 9.81 m/s²

Substituting the values,

7.07 = 2v/9.81
v = 34.68 m/s
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A model rocket fired vertically from the ground ascends with a constant vertical acceleration of 52.7 m/s2 for 1.41 s. its fuel
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For rectilinear motions, derived formulas all based on Newton's laws of motion are formulated. The equation for acceleration is

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We can use v1 for the formula of the maximum height attained by an object thrown upwards:

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The maximum height attained by the model rocket is 281.42 m.

For the amount of time for the whole flight of the model rocket, there are 3 sections to this: time at constant acceleration, time when it lost fuel and reached its maximum height and the time for the free fall.

Time at constant acceleration is given to be 1.41 s. Time when it lost fuel covers the difference of the maximum height and the distance travelled at constant acceleration.

2ax=v2^2-v1^2
2(-52.7)(x) = 0^2-74.31^2
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281.42-52.386 = (0)^2+1/2*(9.81)(t^2)
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Lastly, for free falling objects, the equation is
t = √2y/g = √2(281.42)/9.81 = 7.57 s

Therefore, the total time= 1.41+6.83+7.57 = 15.81 s

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