Question
A ball is projected vertically yowards with a speed of 50ms^-1. On return it passes the point of projection and falls 78m below. Calculate the total time taken
Answers
GPT-4o mini
To solve this problem, we consider the motion of the ball under the influence of gravity. We can break the motion into two parts: the upward motion until the ball reaches its highest point, and the downward motion until it falls 78 m below the point of projection.
### Part 1: Upward Motion
1. **Initial Velocity (\( u \))**: 50 m/s (upward)
2. **Final Velocity at the highest point (\( v \))**: 0 m/s (at the peak)
3. **Acceleration (\( a \))**: -9.81 m/s² (gravity acting downward)
Using the equation of motion:
\[
v = u + at
\]
Setting \( v = 0 \) (at the highest point), we get:
\[
0 = 50 - 9.81t
\]
Solving for \( t \):
\[
9.81t = 50 \\
t = \frac{50}{9.81} \approx 5.1 \text{ seconds}
\]
This is the time taken to reach the highest point.
### Part 2: Downward Motion
Next, we find the time it takes to fall from the highest point down to a point 78 m below the initial launch point.
1. **Initial height at the highest point** \( h \): We first need to calculate this height. Using the formula:
\[
h = \frac{u^2}{2g} = \frac{(50)^2}{2 \times 9.81} \approx \frac{2500}{19.62} \approx 127.4 \text{ meters}
\]
So the total distance fallen will be \( 127.4 + 78 = 205.4 \text{ m} \) downward.
Now we can use the second equation of motion to find the time taken to fall this distance:
\[
s = ut + \frac{1}{2} a t^2
\]
Here,
- \( s = 205.4 \) m (downward)
- \( u = 0 \) (starts falling from rest at the peak)
- \( a = 9.81 \) m/s² (acceleration due to gravity)
Substituting the values into the equation:
\[
205.4 = 0 + \frac{1}{2} \times 9.81 \times t^2
\]
Solving for \( t^2 \):
\[
205.4 = 4.905 t^2 \\
t^2 = \frac{205.4}{4.905} \approx 41.87 \\
t \approx \sqrt{41.87} \approx 6.46 \text{ seconds}
\]
### Total Time
Now, we can find the total time taken for the whole motion (upward and downward):
\[
\text{Total time} = \text{Time up} + \text{Time down} \approx 5.1 + 6.46 \approx 11.56 \text{ seconds}
\]
### Conclusion
The total time taken for the ball to return and fall 78 meters below the point of projection is approximately **11.56 seconds**.
### Part 1: Upward Motion
1. **Initial Velocity (\( u \))**: 50 m/s (upward)
2. **Final Velocity at the highest point (\( v \))**: 0 m/s (at the peak)
3. **Acceleration (\( a \))**: -9.81 m/s² (gravity acting downward)
Using the equation of motion:
\[
v = u + at
\]
Setting \( v = 0 \) (at the highest point), we get:
\[
0 = 50 - 9.81t
\]
Solving for \( t \):
\[
9.81t = 50 \\
t = \frac{50}{9.81} \approx 5.1 \text{ seconds}
\]
This is the time taken to reach the highest point.
### Part 2: Downward Motion
Next, we find the time it takes to fall from the highest point down to a point 78 m below the initial launch point.
1. **Initial height at the highest point** \( h \): We first need to calculate this height. Using the formula:
\[
h = \frac{u^2}{2g} = \frac{(50)^2}{2 \times 9.81} \approx \frac{2500}{19.62} \approx 127.4 \text{ meters}
\]
So the total distance fallen will be \( 127.4 + 78 = 205.4 \text{ m} \) downward.
Now we can use the second equation of motion to find the time taken to fall this distance:
\[
s = ut + \frac{1}{2} a t^2
\]
Here,
- \( s = 205.4 \) m (downward)
- \( u = 0 \) (starts falling from rest at the peak)
- \( a = 9.81 \) m/s² (acceleration due to gravity)
Substituting the values into the equation:
\[
205.4 = 0 + \frac{1}{2} \times 9.81 \times t^2
\]
Solving for \( t^2 \):
\[
205.4 = 4.905 t^2 \\
t^2 = \frac{205.4}{4.905} \approx 41.87 \\
t \approx \sqrt{41.87} \approx 6.46 \text{ seconds}
\]
### Total Time
Now, we can find the total time taken for the whole motion (upward and downward):
\[
\text{Total time} = \text{Time up} + \text{Time down} \approx 5.1 + 6.46 \approx 11.56 \text{ seconds}
\]
### Conclusion
The total time taken for the ball to return and fall 78 meters below the point of projection is approximately **11.56 seconds**.