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Art [367]
2 years ago
11

A rock is thrown downward from an unknown height above the ground with an initial speed of 6.1 m/s. It strikes the ground 1.7 s

later. Determine the initial height of the rock above the ground. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m
Physics
1 answer:
insens350 [35]2 years ago
7 0

Answer:

24.531 m

Explanation:

t = Time taken = 1.7 s

u = Initial velocity = 6.1 m/s

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=6.1\times 1.7+\dfrac{1}{2}\times 9.8\times 1.7^2\\\Rightarrow s=24.531\ m

The initial height of the rock above the ground is 24.531 m

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The answer is  24.84kJ.

We apply the expression for the work done by the heat engine is,

W=E_{i n}-E_{o u t}. Putting all given values in the equation we get the final answer.

What is heat engine?

  • A heat engine is a machine that uses heat to generate power. It draws heat from a reservoir, uses that heat to produce work, such as move a piston or lift weights, and then releases that heat energy into the sink.
  • We are given:The heat input is $E_{i n}=42.53 \mathrm{~kJ}$. The heat output is $E_{o u t}=17.69 \mathrm{~kJ}$.
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