Question

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?

Answer 1

h = vₒ•t - g•t²/2

Answer 2

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)

Answer 3

James. It is exactly the same equation.

1/2 (a)= a/2

Answer 4

Well, then I made a calculation error because the two equation gave me a different answer.