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Softa [21]
4 years ago
9

A coin dropped from the top of a precipice (cliff) takes 2.42 to hit the ground. How highis the precipice (cliff)? .

Physics
1 answer:
goldfiish [28.3K]4 years ago
4 0

Answer:

28.7 m.

Explanation:

Hello,

In this case, since the coin is dropped from the rest condition at which the initial velocity is 0 m/s, we can compute the height of the precipice via the following equation:

y=y_0+v_0*t-\frac{1}{2}gt^2

Whereas the final height is 0, the initial one is to be computed and the gravity is considered as 9.8 m/s², therefore the height turns out:

y=y_0+v_0*t-\frac{1}{2}gt^2\\\\y_0=\frac{1}{2}gt^2=\frac{1}{2}*9.8m/s^2*(2.42s)^2 \\\\y_0=28.7m

It means that the height of the cliff is 28.7 m.

Regards.

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If we put x(t)=d=3.5 km=3500 m into the equation, the corresponding time t is the time it takes for the mass to travel this distance:
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4 years ago
At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After be
max2010maxim [7]

Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

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2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

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

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

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3 years ago
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jeka57 [31]

Explanation:

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T-mg=ma

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Work done by tension, W_t=F\times h

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Explanation:

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