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rjkz [21]
3 years ago
7

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera

from the railing of the basket when it is 18 mm above the ground. If the balloon continues to rise at the same speed, how high is the railing when the camera hits the ground?
Physics
1 answer:
jeka943 years ago
5 0

To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

h = v_0 t -\frac{1}{2} gt^2

-18 = 15*t + \frac{1}{2} 9.8*t^2

t = 3.98s

Then the total distance traveled would be

h = h_0 +v_0t

h = 18+15*3.98

h = 77.7m

Therefore the railing will be at a height of 77.7m when it has touched the ground

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A stationary shell is exploded in to three fragments A, B, C of masses in the ratio 1:2:3. A travels
spin [16.1K]

Answer:

20 m/s

Explanation:

If the mass of fragment A is m, then the mass of fragment B is 2m, and the mass of fragment C is 3m.

The velocity of A is 60 m/s at angle 0°.

The velocity of B is 30 m/s at angle 120°.

The velocity of C is v at angle θ.

In the x direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 cos 0°) + 2m (30 cos 120°) + 3m (v cos θ)

0 = 60m − 30m + 3m v cos θ

0 = 30m + 3m v cos θ

-30m = 3m v cos θ

-10 = v cos θ

In the y direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 sin 0°) + 2m (30 sin 120°) + 3m (v sin θ)

0 = 0 + 30√3 m + 3m v sin θ

-30√3m = 3m v sin θ

-10√3 = v sin θ

Square the two equations and add together:

(-10)² + (-10√3)² = (v cos θ)² + (v sin θ)²

100 + 300 = v² cos² θ + v² sin² θ

400 = v² (cos² θ + sin² θ)

400 = v²

v = 20

The speed of fragment C is 20 m/s.

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Answer:

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Explanation:

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