Question #eb17f

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Question #eb17f

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Answer 1 · 2017-04-19T00:38:36

$20.41 m = h$ $20.41 m= h$

Explanation:

This problem can be solved using the energy equations.

The object is thrown with an initial velocity straight up. We are given a mass, and velocity. The total mechanical energy of the system is only due to KE (energy associated with motion)

$K E = \frac{1}{2} m v^{2}$ $KE = 1/2mv^2$

Where
$m - mass of object (kg)$ $"m - mass of object (kg)"$
$v - velocity of the object (\frac{m}{s})$ $"v - velocity of the object" (m/(s))$

Steps
$K E = \frac{1}{2} m v^{2}$ $KE = 1/2mv^2$
$K E = \frac{1}{2} (0.700 k g) {(20.0 \frac{m}{s})}^{2}$ $KE = 1/2(0.700kg)(20.0m/s)^2$
$K E = 140 k g \frac{m^{2}}{s^{2}}$ $KE = 140 kg m^2/s^2$

At the max height, the mass possesses only PE since at the top, velocity is 0 so KE is 0. So total mechanical energy here is due to PE.

$P E = m g h$ $PE = mgh$

Since energy is conserved and no energy was lost due to outside work, we can set the initial energy of the system (only due to KE) equal to the final energy of the system (only due to PE)

$140 k g \frac{m^{2}}{s^{2}} = m g h$ $140 kg m^2/s^2 = mgh$

$140 k g \frac{m^{2}}{s^{2}} = (0.700 k g) (9.8 \frac{m}{s^{2}}) (h)$ $140 kg m^2/s^2 = (0.700 kg)(9.8 m/s^2)(h)$

$(140 cancel(kg m)^2/cancel(s^2))/((0.700 cancel(kg))(9.8 cancel(m/s^2))) = h$

$20.41 m= h$

$Answer: 20.41 m$

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Question #eb17f

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