Question

a ball is thrown up so that it reaches a height of 25m. how fast was it going when it was 5m high. take g=9.8m/s^2. use the equation of the motion

Answer 1

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:
1. \( a = -g = -9.8 \, \text{m/s}^2 \)
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**.