In this problem, you will apply kinematic equations to a jumping flea. Take the

Question

In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/s2 . Ignore air resistance. Part A A flea jumps straight up to a maximum height of 0.380m . What is its initial velocity v0 as it leaves the ground? Express your answer in meters per second to three significant figures. In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/s2 . Ignore air resistance. Part A A flea jumps straight up to a maximum height of 0.380m . What is its initial velocity v0 as it leaves the ground? Express your answer in meters per second to three significant figures. In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/s2 . Ignore air resistance. In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/s2 . Ignore air resistance. In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/s2 . Ignore air resistance. Part A A flea jumps straight up to a maximum height of 0.380m . What is its initial velocity v0 as it leaves the ground? Express your answer in meters per second to three significant figures. Part A A flea jumps straight up to a maximum height of 0.380m . What is its initial velocity v0 as it leaves the ground? Express your answer in meters per second to three significant figures. Part A A flea jumps straight up to a maximum height of 0.380m . What is its initial velocity v0 as it leaves the ground? Express your answer in meters per second to three significant figures.

Explanation / Answer

Haha, this isn't difficult, it's basic kinematics. There are already equations written to solve exactly this kind of problem. Here is one method of solution.

a) E = mgh = m * 9.80 * 0.380 (gravitational potential energy, GPE)
Since no energy is lost, initial kinetic energy (KE) = GPE. Therefore:
(1/2) * m * v^2 = m * 9.80 * 0.380
Cancel mass from both sides and rearrange to give:
v = sqrt(9.80 * 0.380 * 2) = 1.189m/s

b) a = 9.80
t = 2 * (1.189/ 9.80) = 0.242 s

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