Answer:

To calculate the maximum height reached by the ball, you can use the following formula for projectile motion:

\[H = \frac{V_i^2 \sin^2(\theta)}{2g}\]

Where:

- \(H\) is the maximum height.

- \(V_i\) is the initial velocity (20 m/s).

- \(\theta\) is the angle above the horizontal (45 degrees, which is \(\frac{\pi}{4}\) radians).

- \(g\) is the acceleration due to gravity (9.8 m/s²).

Now, plug in the values and calculate:

\[H = \frac{(20 \, \text{m/s})^2 \cdot \sin^2\left(\frac{\pi}{4}\right)}{2 \cdot 9.8 \, \text{m/s²}}\]

First, calculate \(\sin^2\left(\frac{\pi}{4}\right)\):

\(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)

\(\sin^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}\)

Now, plug this value into the formula:

\[H = \frac{(20 \, \text{m/s})^2 \cdot \frac{1}{2}}{2 \cdot 9.8 \, \text{m/s²}}\]

Calculate:

\[H = \frac{200 \, \text{m²/s²}}{19.6 \, \text{m/s²}}\]

Now, divide to find \(H\):

\[H \approx 10.2 \, \text{meters}\]

Rounding to the nearest meter, the maximum height reached by the ball is approximately 10 meters.

So, the answer is A) 10 m.

Answer:a)10m i think soo without using g=9.8