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hodyreva [135]
3 years ago
7

A package is dropped from a helicopter that is moving upward at15m/s. If it takes 8.0 sec before the package strikes the ground,

how high above the ground was the package when it was released? Neglect air resistance.
Physics
1 answer:
Rainbow [258]3 years ago
7 0

Answer:

313.6 m

Explanation:

From the question given above, the following data were obtained:

Time (t) = 8 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

The height at which the package was dropped can be obtained as follow:

h = ½gt²

h = ½ × 9.8 × 8²

h = 4.9 × 64

h = 313.6 m

Thus, the package was at a height of 313.6 m when it was dropped.

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Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
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        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

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3 years ago
TRUE OR FALSE: The following drops were most likely dropped from a 90 degree angle.
horsena [70]

Answer:

True

Explanation:

If it weren't from a 90 degree angle then the circle would be a bit more oval shaped

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LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

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