Question

Ex 2) A cannon ball is shot straight up into the air with an initial velocity of 25 m/s[Up).

What is the maximum height of the cannonball?

Answer 1

Step-by-step explanation:

The final velocity at the cannon ball's maximum height is zero (
`v_y = 0`). We can use the equation

`v_y^2 = 0 = v_(0y)^2 - 2gy_(max)`

`\Rightarrow y_(max) = (v_(0y)^2)/(2g)`

`y_(max) = \frac{(25\:\text{m/s})^2}{2(9.8\:\text{m/s}^2)}`

`\:\:\:\:\:\:\:=31.9\:\text{m}`