![To determine how high the cannonball will travel, we can use the formula for the maximum height \( h \) reached by an object launched vertically, which is given by:
\[
h = \frac{v^2}{2g}
\]
where:
- \( v \) is the initial velocity (25 m/s),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
Plugging in the values:
\[
h = \frac{(25 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} = \frac{625}{19.62} \approx 31.89 \, \text{m}
\]
Therefore, the correct answer is **31.89 m*](https://theinternetioimages-f2htc2ggcrath6d7.z01.azurefd.net/$web/askai/e747b6fb-7065-4d26-8c75-9a8fe8d2a367.webp)
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Question: A cannon shoots a 200Kg cannonball straight up into the air. The cannonball leaves the cannon at 25m/s. How high will the cannonball travel? 125,012.67 m 6,377.55 m 31.89 m 62,500 m
To find out how high the cannonball will travel, we can use the principles of physics, specifically kinematic equations for projectile motion. We'll assume there's no air resistance for simplicity. The highest point the cannonball will reach (maximum height) can be found using the following kinematic equation: \[ v^2 = u^2 + 2as \] Where: - \( v \) = final velocity (0 m/s at the maximum height) - \( u \) = initial velocity (25 m/s) - \( a \) = acceleration due to gravity (-9.81 m/s², since it's going upwards against gravity) - \( s \) = maximum height (which we want to find) Rearranging the equation to solve for \( s \): \[ s = \frac{v^2 - u^2}{2a} \] Substituting in the known values: \[ s = \frac{0 - (25)^2}{2 \times -9.81} \] \[ s = \frac{-625}{-19.62} \] \[ s \approx 31.85 \, \text{m} \] The closest value given in your options is: \[ \boxed{31.89 \, \text{m}} \] So, the correct answer is 31.89 meters.
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