Determining Initial Speed and Maximum Height of a Thrown Ball
In this article, we will solve a problem involving a ball that is thrown upward and caught after 3.4 seconds. Using the equations of motion under constant acceleration due to gravity, we will determine the initial speed at which the ball was thrown and the maximum height it reached. This example will help you understand the application of equations of motion in real-world scenarios.
Understanding the Scenario
The ball is thrown upward and caught after 3.4 seconds. The time to reach the highest point is half of the total time, so the time to reach the highest point ({t}_{up}) is (frac{3.4}{2} 1.7) seconds.Using Equations of Motion
[text{Final velocity at the highest point} v u - gt]
[0 u - 9.81 text{ m/s}^2 cdot 1.7 text{s}]
[u 9.81 cdot 1.7]
[u approx 16.677 text{ m/s}]
Thus, the initial velocity ({u}) at which the ball was thrown is approximately 16.68 m/s.
[text{Maximum height} h ut - frac{1}{2}gt^2]
[h 16.677 text{ m/s} cdot 1.7 text{s} - frac{1}{2} cdot 9.81 text{ m/s}^2 cdot (1.7 text{s})^2]
[h approx 28.3199 - 14.1895]
[h approx 14.1304 text{ m}]
Therefore, the maximum height ({h}) that the ball reached is approximately 14.13 meters.
Application and Further Insights
Consider the problem where the total time the ball is in the air is 2.5 seconds. This makes the time to reach the maximum height ({t}_{up} 1.25) seconds.
({t}_{up} frac{2.5 text{s}}{2} 1.25 text{s})Using the initial velocity ({v}_{in} 16.677 text{ m/s}), we can find the maximum height.
[1.25 text{s} frac{v}_{up}}{g}]
[v}_{up} 1.25 text{s} cdot 9.8 text{ m/s}^2 12.25 text{ m/s}]
[d_{max} frac{v}_{up}}{2} cdot t_{up}]
[d_{max} frac{12.25 text{ m/s}}{2} cdot 1.25 text{s} 7.66 text{ m}]
Alternatively, using the kinematic equation for constant acceleration, we can determine the initial velocity ({v}_o) and the maximum height ({y}).
[v_f v_o - g cdot t]
[0 v_o - 9.8 cdot 1.25]
[v_o 9.8 cdot 1.25 12.25 text{ m/s}]
[v_f^2 v_o^2 - 2 cdot g cdot y]
[0^2 (12.25 text{ m/s})^2 - 2 cdot (-9.8 text{ m/s}^2) cdot y]
[12.25^2 19.6 cdot y]
[y frac{12.25^2}{19.6} approx 7.65625 text{ m}]
[approx 7.7 text{ m}]
Thus, the maximum height from the ground is approximately 7.7 meters.
Conclusion
This detailed solution demonstrates the application of the equations of motion to solve problems involving the motion of an object under constant acceleration due to gravity. Understanding these principles is crucial in various fields, including physics, engineering, and sports science.