To solve the problem of finding the speed of the ball when it is 5 meters high, we can use the principles of energy conservation or kinematic equations. Here, we'll use the kinematic equations of motion.
Given:
- Initial height \( h_0 = 25 \) m (maximum height)
- Height at which we need to find the speed \( h = 5 \) m
- Acceleration due to gravity \( g = 9.8 , \text{m/s}^2 \)
Using the Kinematic Equation:
We can use the following kinematic equation which relates initial velocity, final velocity, acceleration, and displacement:
\[ v^2 = u^2 + 2a s \]
Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which will be negative since it's going against gravity),
- \( s \) is the displacement.
To find displacement \( s \):
The displacement from the point of release to the height of 5 m is: \[ s = h_0 - h = 25 - 5 = 20 , \text{m} \]
Setting up the equation:
- \( a = -g = -9.8 , \text{m/s}^2 \)
- We need to rearrange the equation to find \( v \):
\[ v^2 = u^2 + 2(-g)(s) \]
Substituting \( s = 20 \) m and \( g = 9.8 , \text{m/s}^2 \):
\[ v^2 = u^2 - 2(9.8)(20) \] \[ v^2 = u^2 - 392 \]
Finding initial velocity \( u \):
To find \( u \), we need to determine the speed at maximum height. At maximum height, the velocity \( v = 0 \). Using the equation from maximum height to \( h = 25 \) m (using \( s = 25 \) m):
\[ 0 = u^2 - 2(9.8)(25) \] \[ u^2 = 2(9.8)(25) \] \[ u^2 = 490 \] \[ u = \sqrt{490} \approx 22.14 , \text{m/s} \]
Now substitute \( u \) back into the equation for \( v \):
\[ v^2 = 490 - 392 \] \[ v^2 = 98 \] \[ v = \sqrt{98} \approx 9.9 , \text{m/s} \]
Conclusion:
The speed of the ball when it is 5 meters high is approximately 9.9 m/s.
1 answer