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baherus [9]
3 years ago
10

A hot-air balloon is rising straight up with a speed of 4.6 m/s. a ballast bag is released from rest relative to the balloon at

13.4 m above the ground. how much time elapses before the ballast bag hits the ground?
Physics
1 answer:
kherson [118]3 years ago
4 0

The time taken by the ballast bag to reach the ground is 2.18 s

The ballast bag at rest with respect to the balloon has the upward velocity (u) of 4.6 m/s , which is the velocity of the balloon. When it is dropped from the balloon, its motion is similar to an object thrown upwards with an initial velocity <em>u </em>and it falls under the acceleration due to gravity<em> g.</em>

Taking the upward direction as positive and the downward direction as negative, the following equation of motion may be used.

s=ut+\frac{1}{2}at^2

The bag makes a net displacement <em>s</em> of 13.4 m downwards, hence

s=-13.4 m

Its initial velocity is

u=+4.6 m/s

The acceleration due to gravity acts downwards and hence it is negative.

g=-9.8 m/s^2

Use the values in the equation of motion and write an equation for t.

s=ut+\frac{1}{2} at^2 \\ -13.4=4.6t-\frac{1}{2}(9.8)t^2\\ 4.9t^2-4.6t-13.4=0

Solving the equation for t and taking only the positive value for t,

t=2.18 s

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3 years ago
block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally
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v_{2}=\frac{1}{8}*(2+2)=\frac{1}{2}

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