You throw a 3.00-N rock vertically into the air from ground level...

Question

You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work–energy theorem to find (a) the rock’s speed just as it left the ground

Accepted Answer

Hey everyone today, we're dealing with the problem about the work energy theorem. So we're being told that a two newton ball, Let's highlight that a two Newton ball is thrown from the ground upwards into the air At a height of 10 m above the ground. The speed of the ball or the velocity is measured as 20 m/s. With this, we're being asked to calculate the initial speed of the ball. So let's go ahead and write out a few things first. The work energy theorem state set the work in this case due to gravity is equal to the final kinetic energy of the object, minus the initial kinetic energy of the object. And we can recall that kinetic energy is equal to one half, multiplied by mass times. Oops wrong parentheses one half multiplied by the mass times velocity squared. So with this in mind, let's go ahead and rewrite our work energy theorem. So we have the work due to gravity is equal to One half and I'll write this in blue for the final velocity. So one half multiplied by M. V two squared minus the reds and red one half Him. The one squared from here. We can also remember that if we're talking about work due to gravity thin this will also be the same thing as native MGH is equal to. We can factor out the one half from either side or 1/2.m. Multiplied by feet two squared minus V. One squared, dividing both sides by M our M. Or our mass values will cancel out And we will get and multiplying both sides by two, Multiplied by two. To cancel on behalf here to isolate our velocity terms. We get to G h negative is equal to V two squared minus the one square Rearranging for V one because we are calculating the initial speed or the initial velocity we get that V one squared is equal to oops is equal to the two square Plus two gravity or the force of gravity multiplied by the height of the ball or the height that the ball travels to. However, we're looking for the initial velocity at the initial velocity squared, so we simply square root both sides and we get that the initial velocity or V one is equal to is equal to the square root of V two squared Plus two. GH substituting in our values. Now We know that the height traveled or the height that the ball was when it was measured was 10 m. The force of gravity is 9.8 m per second squared. That is acceleration due to gravity. My bad not force due to gravity and we know that the final velocity Or the velocity when it was measured was 20 m/s. So substituting in all of these values, we get that it is the square root of 20 meters per second squared Plus two into 9.8 m/s squared Multiplied by the height of 10 m Which equates to a value of 24. Let me write this in green 24. m/s, therefore the initial velocity of the ball or initial speed, according to the work energy theorem. His answer choice B 24.4 m per second. I hope this helps, and I look forward to seeing you all in the next one.

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