A tennis ball on Mars, where the acceleration due to gravity is 0...

Question

A tennis ball on Mars, where the acceleration due to gravity is 0.379g and air resistance is negligible, is hit directly upward and returns to the same level 8.5 s later. (a) How high above its original point did the ball go?

Accepted Answer

Hello everyone. So in this problem of football on Earth where the acceleration due to gravity is G is kicked vertically upward by a player so that it returns to its starting point after 10 seconds. What is the maximum height reached by the football? So we have some football player sit up with some initial velocity undergoes freefall, it reaches this maximum height. H max is what we are looking for and the total time Is equal to 10 seconds. You also recall that the time from the starting point to the H max is just half of the total time and it's the same from h max to back to its starting point. T is equal to five seconds. These are just half trips and it falls back down Which is its starting point. And throughout this free fall the acceleration is simply the acceleration due to gravity negative 9.81 m/s square. And it's also important to realize that at this age max the final velocity or the velocity in general actually is equal to zero meters for a second. So we can start off by recalling the cinematic equations we could use to solve this problem. So the F one Is equal to VI plus 80. The second one, V F squared is equal to V I squared plus two A delta X. The 3rd 1 Delta X is equal to Vitt plus 1.5 a T squared. Now looking at these cinematic equations, we want to solve H max and at this the delta H for both of these half trips. also just got the ax as both of them are just a measure of distance is equal to h max. And the only two equations that has a church or some form of distance, is equation to an equation three. And we are given V i And v equals zero m/s. We are given time in the second phase or second half trip from each max to back to the zero. And we also know that it is in free fall. So we can simply use this third Kinnah Matic equation to solve for H max. Rewrite it as H max is equal to V I T plus one, a half a t squared. We know that the eye is equal to zero in this second phase. And now we can simply solve for age max, so then acceleration due to gravity And so then the half time five seconds squared and we simply get that h max, It was equal to 122.5 m, which is answer choice B of this house have a great day

A tennis ball on Mars, where the acceleration due to gravity is 0.379g and air resistance is negligible, is hit directly upward and returns to the same level 8.5 s later. (a) How high above its original point did the ball go?

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