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. (a) What is the initial speed of the egg?

Video transcript

Hello everyone in this problem, A person kicks a football vertically upward from the edge of a cliff of height 50 m. The football passes him on its way down and passes a 500.40 m below the point where the football left the foot of the person six seconds after it was thrown. And air resistance is negligible. Then what is the initial speed of the football? So the first step is to probably draw a picture which would help visualize the problem. So we have a person on some cliff 50 m above some age equals zero reference point Since that age is 50 m and he throws some ball with an initial velocity of V. I, which is what we're trying to find. And as it's thrown it undergoes freefall and reaches some point with a difference from the initial to the final Delta H is equal to negative 40 m. And we recall that under free fall an object and objects acceleration is simply just the acceleration due to gravity. So a. Y. Is equal to G. Which is equal to negative 9.81 m per second squared. And from the problem statement were also given that the time in free fall is six seconds, And we also know that Delta H is -40 m. Recall the cinematic equations where the first one is the F is equal to V. I plus 80 and the second one is V. F squared is equal to V I squared Plus two. A Delta X. And the third one is delta X is equal to V. I. T plus one half a T squared. So now to determine which kind of that equation we could use we could just look at what values we are given were given delta H. Were given T. And we're also given some acceleration G. And as you can see we're trying to find V. I. So the equipped beach equation has some V. I. But only equation three has the values were given to the X. Which is the age, it's just some distance T. And we're also given A which is the acceleration due to gravity. So we could use this third cinematic equation to solve this. Rewrite this as delta H is equal to V I. T plus one half G. T squared. You can rearrange it for V. I. Now you get the negative V. I. T. Is equal to negative delta H plus one half gt squared V. I. Is equal to delta H minus one half Gt squared over T. And now we can simply plug in the values Delta H. As we recall is negative 40 m minus one half G is negative 9. meters per second squared. He was six seconds. Also divide this whole thing by six seconds, Let me get that v. I. is equal to 22.7 m/s, which is answer choice B 22.7 m/s. Have a great day. Everyone