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IrinaK [193]
2 years ago
14

Batman (mass = 96.1 kg) jumps straight down from a bridge into a boat (mass = 458 kg) in which a criminal is fleeing. The veloci

ty of the boat is initially +11.3 m/s. What is the velocity of the boat after Batman lands in it?
Physics
1 answer:
MariettaO [177]2 years ago
8 0

Answer:

The velocity of the boat after the batman lands in it is +9.26 m/s

Explanation:

Applying the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Note: The collision between the Batman and the boat is an inelastic collision.

m'u'+mu = V(m+m').................... Equation 1

Where m' = mass of the Batman, u' = initial velcoity of the batman, m = mass of the boat, u = initial velocity of the boat, V = common velocity.

make V the subject of equation 1

V = (m'u'+mu)/(m+m')............... Equation 2

Given: m' = 96.1 kg, u' = 0 m/s, m = 458 kg, u = +11.3 m/s.

Substitute these values into equation 2

V = [(96.1×0)+(458×11.2)]/(96.1+458)

V = 5129.6/554.1

V = +9.26 m/s

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Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

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1 year ago
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Answer:

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Explanation:

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F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

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∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

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