Question
A person throws a ball upward into the air with an initial velocity of 11 m/s. How high will it go before it comes to rest?
Answers
GPT-4o mini
To find out how high the ball will go before it comes to rest, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement. The relevant kinematic equation is:
\[
v^2 = u^2 + 2a s
\]
where:
- \( v \) is the final velocity (0 m/s at the highest point),
- \( u \) is the initial velocity (11 m/s),
- \( a \) is the acceleration (which will be -9.81 m/s² due to gravity, acting downward), and
- \( s \) is the displacement (the height the ball reaches).
Plugging in the values:
\[
0 = (11)^2 + 2(-9.81)s
\]
Solving for \( s \):
1. Calculate \( (11)^2 \):
\[
121 = 2(-9.81)s
\]
2. If we simplify the equation:
\[
121 = -19.62s
\]
3. Solve for \( s \):
\[
s = \frac{121}{19.62}
\]
Calculating this:
\[
s \approx 6.17 \text{ meters}
\]
Thus, the maximum height the ball will reach before it comes to rest is approximately **6.17 meters**.
\[
v^2 = u^2 + 2a s
\]
where:
- \( v \) is the final velocity (0 m/s at the highest point),
- \( u \) is the initial velocity (11 m/s),
- \( a \) is the acceleration (which will be -9.81 m/s² due to gravity, acting downward), and
- \( s \) is the displacement (the height the ball reaches).
Plugging in the values:
\[
0 = (11)^2 + 2(-9.81)s
\]
Solving for \( s \):
1. Calculate \( (11)^2 \):
\[
121 = 2(-9.81)s
\]
2. If we simplify the equation:
\[
121 = -19.62s
\]
3. Solve for \( s \):
\[
s = \frac{121}{19.62}
\]
Calculating this:
\[
s \approx 6.17 \text{ meters}
\]
Thus, the maximum height the ball will reach before it comes to rest is approximately **6.17 meters**.