A stone is thrown vertically upward with a speed of 25.0 m/s. a. How fast is it

Question

A stone is thrown vertically upward with a speed of 25.0 m/s.
a. How fast is it moving when it reaches a height of 12.0 m? in m/s
b. How long is required to reach this height? in secs
c.Why are there two answers to part B?
explain part c in paragraph form. less than 500 words.

Explanation / Answer

(a) U=25 m/sec V^2 =U^2-2*g*h V^2=625-2*9.8*12 => V=19.74m/sec (b) s=U*t -0.5*g*t^2 => 12=25*t-0.5*9.8*t^2=> 4.9*t^2 - 25*t+12=0 t1=0.5363 sec t2=4.5656 sec (c) Part b has two answers because ball after reaching maximum height returns back and again passes from the same point so it crosses same point at two different times

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