A stone is thrown vertically upward with a velocity of 19.6 m/s. After...
Problem Statement:
A stone is thrown vertically upward with a velocity of 19.6 m/s. After 2 seconds, another stone is thrown upwards with a velocity of 9.8 m/s. When and where will these stones collide? Explain in details.
Solution:
1. Finding the equation for the position of the stones:
For each stone, we can find the equation for its position as a function of time using the following formula:
Position = Initial Position + Initial Velocity * Time + 0.5 * Acceleration * Time^2
Since the stones are thrown vertically upward, the initial position is zero.
For the first stone, the initial velocity is 19.6 m/s and the acceleration is -9.8 m/s^2 (negative due to gravity pulling the stone downwards).
Therefore, the equation for the position of the first stone, y1, as a function of time, t, is:
y1(t) = 19.6t - 4.9t^2
For the second stone, the initial velocity is 9.8 m/s and the acceleration is also -9.8 m/s^2.
Therefore, the equation for the position of the second stone, y2, as a function of time, t, is:
y2(t) = 9.8(t-2) - 4.9(t-2)^2
2. Finding the time and position of collision:
Since the two stones will collide when their positions are equal, we can set y1(t) equal to y2(t) and solve for t:
19.6t - 4.9t^2 = 9.8(t-2) - 4.9(t-2)^2
Expanding and simplifying, we get:
-4.9t^2 + 19.6t - 19.6 = 0
Dividing both sides by -4.9, we get:
t^2 - 4t + 4 = 0
Using the quadratic formula, we get:
t = 2 ± sqrt(0)
Therefore, t = 2.
Substituting t = 2 into either of the position equations, we get:
y1(2) = 19.6(2) - 4.9(2)^2 = 19.6 m
Therefore, the stones will collide at a height of 19.6 meters above the ground.
3. Conclusion:
The stones will collide after 2 seconds at a height of 19.6 meters above the ground. This was found by setting the position equations for each stone equal to each other and solving for the time of collision. The position of collision was then found by substituting the time of collision into either of