# A ball is thrown into the air from the ground. The ball reaches a max height of 8m above the ground at 3 seconds. ?

Find an equation to model this situation

### 1 Answer

- NCSLv 72 months ago
In general, y = y₀ + v₀*sinΘ*t + ½at²

where v₀ is the launch velocity and Θ the angle from horizontal.

Here we have y₀ = 0 and a = -9.8 m/s², so

y = v₀*sinΘ*t - 4.9m/s²*t²

If max height is achieved in 3 seconds, then the time of flight is 6 seconds.

Time of flight t = 2·V·sinΘ/g = 6 s

V*sinΘ = 6s * 9.8m/s² / 2 = 27.4 m/s

max. height h = (V·sinΘ)² / (2g) = (27.4m/s)² / 2*9.8m/s² = 38 m

But this does not equal the "8 m" you gave in the problem statement. So unless this experiment took place somewhere other than the surface of the Earth, the data is inconsistent.

But IF you meant "a max height of 38 m...", then

y = 27.4m/s*t - 4.9m/s²*t²

will give you the vertical motion of the ball.

The x-motion cannot be modeled without more data.