Calculus: A ball is thrown vertically upwards at 20ms^-1 against the constant acceleration.. (cont)?

v(t) = Vo + at

v(t) = 20 -10t

h(t) = Vot + 1/2at^2

h(t) = 20t - 5t^2

The ball returns to its starting point when the final velocity (Vt) equals to - initial velocity (Vo), so:

Vt = - Vo

Using this equation:

Vt = Vo + at

- Vo = Vo + at

- 2Vo = at

-40 = -10t

t = 4s

Or, to get physical insight on this problem, we know that at the highest height the ball travels, the velocity of the ball will equals to 0, so:

Vt = Vo + at

0 = 20 - 10t

10t = 20

t = 2s

The ball will travel down at the same acceleration, so at the same point where it was thrown, the velocity will have the same magnitude but differ in direction.

So, just multiply the time by 2, we get:

2t = 2x2 = 4s

V^2= U^2 + 2gS here S= 0 because of the fact the ball is coming back to an identical factor so preliminary velocity equivalent to velocity that hit the floor in user-friendly terms the direction differs so taking upward direction answer is -7.8ms-a million