Starting from rest, a 63.0 kg woman jumps down to the floor from a height of 0.660 m and immediately jumps back up into the air. While she is in contact with the ground during the time interval \(0 < t < 0.800 \, \text{s}\), the force she exerts on the floor can be modeled using the function:\[F = 9,200t - 11,500t^2\]where \(F\) is in newtons and \(t\) is in seconds.**(a)** What impulse (in N · s) did the woman receive from the floor? (Enter the magnitude. Round your answer to at least three significant figures.)\[ \text{Impulse} = \boxed{} \, \text{N · s} \]**(b)** With what speed (in m/s) did she reach the floor? (Round your answer to at least three significant figures.)\[ \text{Speed when reaching floor} = \boxed{} \, \text{m/s} \]**(c)** With what speed (in m/s) did she leave it? (Round your answer to at least three significant figures.)\[ \text{Speed when leaving floor} = \boxed{} \, \text{m/s} \]**(d)** To what height (in m) did she jump upon leaving the floor?\[ \text{Jump height} = \boxed{} \, \text{m} \]

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Description: Starting from rest, a 63.0 kg woman jumps down to the floor from a height of 0.660 m and immediately jumps back up into the air. While she is in contact with the ground during the time interval \(0 < t < 0.800 \, \text{s}\), the force she exerts on

Hai, since there are multiple sub parts posted, we will answer the first three sub parts. (a) The impulse received in the time interval ti<t<tf is given by, I=∫titfFdt Substituting, I=∫00.8009200t-11500t2dt=9200t22-11500t3300.800=92000.80022-115000.80033-0=981.3 N·s (b) According to the principle of conservation of energy, the total energy of the woman must be conserved. Initially since the woman starts from rest, her initial kinetic energy will be zero. The total initial energy of the woman will be equal to her initial potential energy. When the woman reaches the floor, her potential energy will be zero. The energy conservation gives, mgh=12mv2 Here, m is the mass of the woman, g is the acceleration due to gravity, h is the initial height of the woman and v is the speed with which the woman reaches the floor. Rearranging the equation, gh=12v2v=2gh=29.80 m/s20.660 m=3.597 m/s(c) According to the impulse-momentum theorem, the impulse received must be equal to the change in momentum of the woman. It is given by, I=mv'-v Here, v' is the speed with which the woman leaves the floor. Rearranging the equation, I=mv'-mvv'=I+mvm=981.3 N·s+63.0 kg3.597 m/s63.0 kg=19.17 m/s

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Starting from rest, a 63.0 kg woman jumps down to the floor from a helght of 0.660 m, and Immediately jumps back up Into the alr. While she is in contact with the ground during the time Interval 0

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Newton's Second Law

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Starting from rest, a 63.0 kg woman jumps down to the floor from a height of 0.660 m and immediately jumps back up into the air. While she is in contact with the ground during the time interval \(0 < t < 0.800 \, \text{s}\), the force she exerts on the floor can be modeled using the function:

\[
F = 9,200t - 11,500t^2
\]

where \(F\) is in newtons and \(t\) is in seconds.

**(a)** What impulse (in N · s) did the woman receive from the floor? (Enter the magnitude. Round your answer to at least three significant figures.)

\[ \text{Impulse} = \boxed{} \, \text{N · s} \]

**(b)** With what speed (in m/s) did she reach the floor? (Round your answer to at least three significant figures.)

\[ \text{Speed when reaching floor} = \boxed{} \, \text{m/s} \]

**(c)** With what speed (in m/s) did she leave it? (Round your answer to at least three significant figures.)

\[ \text{Speed when leaving floor} = \boxed{} \, \text{m/s} \]

**(d)** To what height (in m) did she jump upon leaving the floor?

\[ \text{Jump height} = \boxed{} \, \text{m} \]

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