Asked by James

An object is thrown upward with a speed of 21.5 m/s. How high above the projection point is it after 2.06 s?
Answered by Elena
h = vₒ•t - g•t²/2
Answered by James
Thank you. This was really helpful. I was never given this equation.

The only equation I got was:

y= y_o + v_yo - 1/2(gt^2)
Answered by bobpursley
James. It is exactly the same equation.

1/2 (a)= a/2
Answered by James
Well, then I made a calculation error because the two equation gave me a different answer.
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