Two stones are thrown vertically upwards with the same velocity of 49m/s. If they are thrown one after the other with 3 seconds in between, what is the height at which they collide?
1st: 49t-4.9t^2
2nd: 49(t-3)-4.9(t-3)^2
So, when are the heights equal?
49t-4.9t^2 = 49(t-3)-4.9(t-3)^2
t=6.5
Now just plug that in to find the height.
To determine the height at which the two stones collide, we need to analyze the motion of each stone independently and find the point where their paths intersect.
Let's break down the problem step by step:
1. Calculate the motion of the first stone:
- Initial velocity (u): 49 m/s (upward).
- Time of flight (t): This stone continues to move upwards until it comes to rest and reverses direction. At the highest point, its velocity becomes zero. Since its acceleration due to gravity (g) is -9.8 m/s^2, the time of flight can be calculated using the formula:
t = u / g
t = 49 m/s / 9.8 m/s^2 = 5 seconds
- Maximum height reached by the first stone: Using the equation of motion:
h = ut + (1/2)gt^2
h = (49 m/s)(5 s) + (1/2)(-9.8 m/s^2)(5 s)^2
h = 245 m - 122.5 m = 122.5 m
2. Calculate the motion of the second stone:
- As mentioned, the second stone is thrown 3 seconds after the first stone.
- Therefore, the total time taken by the second stone will be 5 + 3 = 8 seconds.
- Since the time of flight of the second stone is longer, it will reach a greater height.
- Using the same equation of motion, we can find the maximum height reached by the second stone:
h = ut + (1/2)gt^2
h = (49 m/s)(8 s) + (1/2)(-9.8 m/s^2)(8 s)^2
h = 392 m - 313.6 m = 78.4 m
3. Calculate the height at which the two stones collide:
- The collision occurs when both stones are at the same height.
- To find this height, we subtract the height reached by the first stone from the height reached by the second stone:
h_collision = h_second_stone - h_first_stone
h_collision = 78.4 m - 122.5 m
h_collision ≈ -44.1 m
Note: A negative value for height indicates that the collision occurs below the starting point of the stones.