![To find out how high the cannonball will travel, we can use the kinematic equation for objects in free fall:
\[ h = \frac{v^2}{2g} \]
where:
- \( h \) is the maximum height,
- \( v \) is the initial velocity (25 m/s),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \)).
Plugging in the values:
\[ h = \frac{(25 \, m/s)^2}{2 \times 9.81 \, m/s^2} \]
\[ h = \frac{625}{19.62} \]
\[ h \approx 31.89 \, m \]
So the answer is **a. 31.89 m**.](https://theinternetioimages-f2htc2ggcrath6d7.z01.azurefd.net/$web/askai/87e511e9-5cb9-42b3-a18e-feb76f3cf3b6.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? a.31.89 m b.125,012.67 m c.62,500 m d.6,377.55 m
To determine how high the cannonball will travel, we can use the principle of conservation of energy or the kinematic equations. In this case, we'll use the kinematic equation for vertical motion under gravity. The kinematic equation we can use is: \[ v^2 = u^2 + 2as \] where: - \( v \) is the final velocity (0 m/s at the maximum height) - \( u \) is the initial velocity (25 m/s) - \( a \) is the acceleration due to gravity (-9.81 m/s², negative because it's acting downward) - \( s \) is the displacement (height in this case) Rearranging the equation to solve for \( s \): \[ s = \frac{v^2 - u^2}{2a} \] Substitute the values: \[ s = \frac{0^2 - (25)^2}{2 \times -9.81} \] \[ s = \frac{-625}{-19.62} \] \[ s = 31.88 \text{ (rounded to 2 decimal places)} \] So, the height the cannonball will travel is approximately 31.88 meters. Therefore, the correct answer is: a. 31.89 m
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