An egg is thrown nearly vertically upward from a point near the c...

Question

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on the way down and passes a point 30.0 m below its starting point 5.00 s after it leaves the thrower's hand. Ignore air resistance. (b) How high does it rise above its starting point?

Accepted Answer

Welcome back everybody. We have a man sitting on top of a cliff and he throws a ball directly upward In which then it comes right back down and passes him on the way down. Now this cliff is said to be m off the ground but we are told that at some point 40 m below where it originally started It passes and it only takes six seconds to do that. And we are tasked with finding what the maximum height is for this fall. Now, in order to find the maximum height, we are going to need to know the initial velocity at which the ball was thrown upwards. So let's use a Kinnah Matic equation to find this value. First we have the equation that the final vertical position minus the initial vertical position is equal to our initial vertical velocity times time plus 1/2 times our acceleration T where now we are going to be looking At the journey that it takes from when it's originally thrown to the time of six seconds when it is 40 m below. That allows us to fill in T. R. acceleration which is just the acceleration due to gravity and our y minus why not? So let's go ahead and do that or why minus why not? Well, we have 40 m below where it started. So the difference is negative 40. This is equal to our initial velocity. That's our time. six seconds plus one half times the acceleration which like I said, it's just the acceleration due to gravity or negative 9. and then this is times our time square. Now subtracting this term from both sides and then divided by six. We get that our initial velocity is equal to negative 40 plus since we have a negative here and a negative here, 1/2 times 9. times six squared all over six. Need to plug this into your calculator. We get that. Our initial velocity is 22.7 m per second. So we have found our initial velocity now before trying to find a maximum height. A couple more things I want to point out at this top point, our velocity is going to be equal to zero. Now we don't know how much time it takes to reach this top point but we can use our initial velocity, our final velocity and our acceleration along with this little distance that it travels here to find our maximum height. Do this. We are going to use a different cinematic equation. We are going to use that our final vertical velocity squared equal to our initial Vertical velocity squared plus two times our vertical acceleration times are change in distance. Let's go ahead and fill in some values here. We have that our final velocity from here to the top point zero squared This is equal 22.7 squared Plus two times our acceleration due to gravity times are changing distance while we're trying to find the maximum height. So the distance from our starting point to our top point is simply our maximum height max one, subtracting 22.7 squared on both sides. Then dividing By this term. Right here we get that our maximum height equal to negative 22.7 squared all over two times negative 9.81. Which when you plug into your calculator, You get that. Your maximum height is equal to 26.3 m corresponding to our answer choice. A thank you guys so much for watching. Hope this video helped and we will see you all in the next one.

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