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 (b) its maximum height.

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Hey everyone. So today we're using the work energy theorem to sort of calculate the highest point of a balls travel. Now we're told that this to newton ball is thrown from the ground upwards in the air And at a point at a height of 10 m above the ground, rather The speed of the ball is measured as 20 m/s. So we're going to want to use the work energy theorem. So the work energy theorem states that the network of a system is equal to the change in its kinetic energies, which is the same thing as saying the final kinetic energy minus the initial kinetic energy initial. Now, with this in mind and remembering that uh kinetic energy can be defined as one half the mass of an object, multiplied by its velocity squared. We can go ahead and bring that back into our equation, but we can also remember due to the law of conservation of energy, that the change in potential energy is equal to the change in kinetic energy, right? So if we factor that into account and we're dealing with a ball being thrown into the air, it is being thrown directly to a height of m so far At that height of 10 m, were measuring its height or its velocity, my man. But since we're dealing with simply going up and down, we're dealing with gravitational potential energy, it's relying on the force of gravity to bring it back down. So for calling that gravitational potential energy equal to M. G. H. We can substitute that in. And since we're taking the change in uh gravitational potential energy will have negative MG H is equal to one half M. The final squared minus one half. M v not squared. We're not as initial. So we can factor out the right side. So we get MGH is equal to one half M into the final squared minus V, not squared. So r m s will cancel out and we can multiply both sides by two and we get negative two. G H is equal to V. F squared minus v, not squared. And finally solving for V uh initial because we know what the final height is. We know what the final height is. It's 10 m. We need to find the initial height. So we get the initial height is equal to the square root of the final squared plus two G h. And you may be asking, well, why did we do this? Why are we trying to find the initial velocity? Well, we're trying to find the ball's maximum height, but we're not actually given what that height is. And we don't know what the initial velocity of the ball is in order to calculate that either. So with that in mind going back and substituting in our values, we get that. This is equal to Just based purely on the information given here. 1st. In order to solve for the initial velocity, we get that, this is equal to the oops, is equal to the square root of 20 m/s squared plus two into 9.81 meters per second squared Because that is the acceleration due to gravity multiplied by the height 10 m. And that gives us a final answer of 24.4 m/s. So this is our initial velocity. So with all of that in mind, with all that here we are now trying to find the maximum height, right? And what does it mean when we're at maximum height? Well, when we're at maximum height and let's go back to our diagram here and ignoring the fact that we have this maximum height written here. Now we're dealing with maximum height, not just the 10 m that was observed earlier at maximum height. That is when the ball ceases going up and begins its descent and there is a moment in time where the ball has a Vertical velocity of zero. So we now know that the vertical velocity that we're looking for, the final velocity should be zero, which means if we go back to this step in the equation, if we have um excuse me if we have and I'll write this in blue if we have negative MG H is equal to one half m into the final squared minus v initial squared and we're solving for H this time while our masks cancel again and long story short, we get a final answer that H is equal to or a final formula that h the height is therefore equal to. We are the final, oops, the initial squared minus V final squared over two G. And we know that uh the final zero so we can just go ahead and say that this is 24.4 m per second squared zero because again, we've reduced that the at maximum height, the final velocity will be zero divided by 9.81 m/s squared into two. And solving disk is as a final answer of 30.4 meters. Therefore the ball's highest point based on the information in the passage using the work energy theorem is answer choice B 30.4 m. I hope this helps. And I look forward to seeing you all in the next one.

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 (b) its maximum height.

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Key Concepts

Work-Energy Theorem

Kinetic Energy

Potential Energy