A ball is thrown down vertically with an initial speed of 32 ft/s from a height

Question

A ball is thrown down vertically with an initial speed of 32 ft/s from a height of 47 ft. (a) What is its speed (in ft/s) just before it strikes the ground? (b) How long does the ball take to reach the ground? What would be the answers to (c) part a and (d) part b if the ball were thrown upward from the same height and with the same initial speed ? Before solving any equations, decide whether the answers to (c) and (d) should be greater than, less than, or the same as in (a) and (b). Could you please explain in detail. step by step. Thanks

Explanation / Answer

a)v^2-u^2=2*g*h we get v=63.62ft/sec b)47=32t+0.5*32.17*t^2 t=0.98sec c)it is obvious that it takes greater time d)its velocity will be the same...can be explained by energy conservation.. mgh=0.5*m*v^2

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