Determined Trajectories: Calculating Initial Speed and Impact Velocity of an Upward Thrown Stone
In this article, we will delve into the physics behind the motion of a stone thrown vertically upwards. Specifically, we will calculate the initial speed of the stone and its impact velocity when it hits the ground after reaching a certain height. This is a classic problem in kinematics, and understanding these concepts is crucial for anyone studying physics or engineering.
Calculating the Initial Speed of the Stone
Imagine a stone being thrown vertically upwards from a height of 25.6 meters. It reaches a maximum height of 28.8 meters above the ground. The goal is to determine the initial speed of the stone and its velocity upon impact with the ground again. We will break this down into two parts:
Part 1: Rising to Maximum Height
Height of initial throw: 25.6 meters Maximum height: 28.8 meters Acceleration due to gravity: (9.8 , text{m/s}^2)Here's the step-by-step approach to finding the initial speed:
First, calculate the distance the stone travels during its ascent using the formula (s y_{max} - y_o). In this case, (s 28.8 , text{m} - 25.6 , text{m} 3.2 , text{m}). We know that at the maximum height, the final velocity (v_f) is zero. Use the kinematic equation (v_f^2 v_o^2 2 cdot a cdot s). Rearrange the equation to solve for the initial velocity (v_o): [v_o^2 v_f^2 - 2 cdot a cdot s 0 - 2 cdot 9.8 , text{m/s}^2 cdot 3.2 , text{m}] Simplify and solve: [v_o^2 -62.72 , text{m}^2/text{s}^2] [v_o sqrt{62.72} , text{m/s} frac{28sqrt{2}}{5} , text{m/s} approx 7.92 , text{m/s}]Calculating the Final Velocity of the Stone
Now, we need to calculate the final velocity as the stone hits the ground again. This involves a free fall from a height of 28.8 meters. Here’s the step-by-step approach:
Height of the fall: 28.8 meters Initial velocity at the start of this fall: (u 0 , text{m/s}) Acceleration due to gravity: (a 9.8 , text{m/s}^2) Final velocity (v_f) can be calculated using the kinematic equation (v_f^2 u^2 2 cdot a cdot s). Rearrange the equation: [v_f^2 0^2 2 cdot 9.8 , text{m/s}^2 cdot 28.8 , text{m}] Simplify and solve: [v_f^2 564.48 , text{m}^2/text{s}^2] [v_f sqrt{564.48} , text{m/s} frac{84sqrt{2}}{5} , text{m/s} approx 23.76 , text{m/s}]Conclusion
The stone’s initial speed when thrown upwards from a height of 25.6 meters was approximately 7.92 meters per second. As the stone reached the ground again, it struck the ground with a velocity of approximately 23.76 meters per second.
Understanding the calculations and concepts involved in these trajectories is essential for various applications in physics and engineering. Whether you're designing a catapult, analyzing the motion of satellites, or simply solving homework problems, this knowledge can be invaluable.