Question

(4) a ball is thrown 25 cm/s straight up into the air (no air resistance) from ground level. how high will the ball go before it comes back down? (31.9 m)

Explanation:

Step1: Identify the knowns and unknowns

We know the initial velocity \( u = 25 \, \text{m/s} \), the acceleration due to gravity \( a=-g = - 9.8 \, \text{m/s}^2 \) (negative because it acts in the opposite direction of motion), and at the maximum height, the final velocity \( v = 0 \, \text{m/s} \). We need to find the maximum height \( s \).

We use the kinematic equation \( v^{2}=u^{2}+2as \)

Step2: Rearrange the formula to solve for \( s \)

From \( v^{2}=u^{2}+2as \), we can rearrange it to \( s=\frac{v^{2}-u^{2}}{2a} \)

Step3: Substitute the values into the formula

Substitute \( v = 0 \), \( u = 25 \, \text{m/s} \) and \( a=- 9.8 \, \text{m/s}^2 \) into the formula:

\( s=\frac{0-(25)^{2}}{2\times(- 9.8)}=\frac{- 625}{- 19.6}\approx31.9 \, \text{m} \)

Answer:

The ball will rise approximately \( \boldsymbol{31.9 \, \text{m}} \) before it comes back down.