In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assum
1 answer:
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Answer:
The rider's speed will be approximately 35 m/s
Explanation:
Initially the rider has kinetic and potential energy, and after going down the hill, some of the potencial energy turns into kinetic energy. So using the conservation of energy, we have that:
The kinetic and potencial energy are given by:
So we have that:
So the rider's speed will be approximately 35 m/s
Answer: The correct answer is option b.
Explanation: We are given that the rocket is at rest initially final velocity is 445m/s.
The acceleration of the rocket is
To calculate the distance of rocket, we use third equation of motion, which is:
where, v = final velocity = 445m/s
u = initial velocity = 0m/s
a = acceleration =
s = distance = ? m
Putting values in above equation, we get: