Two objects were projected vertically upward at different times with speed of 80m/s and 10m/s respectively,if the time interval is 2s. When and where would they meets
you want
10t - 4.9t^2 = 80(t-2) - 4.9(t-2)^2
solve for t
To determine when and where the two objects meet, we can use the equations of motion. Let's assume the upward direction is positive.
First, we need to calculate the time it takes for each object to reach its maximum height. We can use the equation:
v = u + at
where:
v = final velocity (0 m/s at the top)
u = initial velocity
a = acceleration (in this case, the acceleration due to gravity, which is -9.8 m/s^2)
For the object initially traveling at 80 m/s:
0 = 80 - 9.8t1
9.8t1 = 80
t1 = 80 / 9.8
t1 ≈ 8.16 seconds
For the object initially traveling at 10 m/s:
0 = 10 - 9.8t2
9.8t2 = 10
t2 = 10 / 9.8
t2 ≈ 1.02 seconds
From the information given, we know that the time interval between the two projections is 2 seconds. Therefore, the second object was launched 2 seconds after the first object.
Now, we can find the height that each object reaches using the equation:
s = ut + (1/2)at^2
For the object initially traveling at 80 m/s:
s1 = 80 * t1 + (1/2) * (-9.8) * t1^2
s1 = 80 * 8.16 + (1/2) * (-9.8) * (8.16)^2
s1 ≈ 326.41 meters
For the object initially traveling at 10 m/s:
s2 = 10 * t2 + (1/2) * (-9.8) * t2^2
s2 = 10 * 1.02 + (1/2) * (-9.8) * (1.02)^2
s2 ≈ 5.1 meters
Since the second object was launched 2 seconds after the first object, we need to find out how much time the first object took to reach the height of the second object.
We can use the equation:
s = ut + (1/2)at^2
For the second object (s2 = 5.1 meters):
5.1 = 80 * t3 + (1/2) * (-9.8) * t3^2
Simplifying the equation, we get a quadratic equation:
-4.9t3^2 + 80t3 - 5.1 = 0
Using the quadratic formula, we can solve for t3:
t3 = (-b ± √(b^2 - 4ac)) / 2a
where:
a = -4.9
b = 80
c = -5.1
Solving the equation, we find:
t3 ≈ 0.07 seconds (ignoring the negative root)
Now, to determine when and where the two objects meet, we need to find the total time for the first object:
total time = t1 - t3
total time ≈ 8.16 - 0.07
total time ≈ 8.09 seconds
To find the total distance traveled by the first object:
total distance = 80 * total time + (1/2) * (-9.8) * total time^2
total distance = 80 * 8.09 + (1/2) * (-9.8) * (8.09)^2
total distance ≈ 328.56 meters
Therefore, the two objects meet approximately 8.09 seconds after the first object was launched, at a height of approximately 328.56 meters.
To find out when and where the two objects will meet, we need to analyze their vertical motion.
Let's assume that the objects start from the same initial height and meet at a certain height above the starting point.
We can start by finding the time it takes for each object to reach its highest point. The time taken to reach the highest point can be found using the formula:
t = (v - u) / a
Where:
t is the time taken,
v is the final velocity (0 m/s at the highest point),
u is the initial velocity, and
a is the acceleration due to gravity (approximately -9.8 m/s^2 in the upward direction).
For the first object:
Initial velocity (u1) = 80 m/s
Final velocity (v1) = 0 m/s
t1 = (0 - 80) / (-9.8)
t1 = 8.16 seconds (approximately)
For the second object:
Initial velocity (u2) = 10 m/s
Final velocity (v2) = 0 m/s
t2 = (0 - 10) / (-9.8)
t2 = 1.02 seconds (approximately)
Since the time interval between their projections is 2 seconds, we can calculate the time taken by the second object to meet the first object after being projected. This can be found as follows:
Time taken by the second object = t1 - t2
Time taken by the second object = 8.16 - 1.02
Time taken by the second object = 7.14 seconds (approximately)
Now, we can calculate the distance covered by each object during this time:
Distance covered by the first object:
S1 = u1 * t1 + (1/2) * a * t1^2
S1 = 80 * 8.16 + (1/2) * (-9.8) * (8.16)^2
S1 ≈ 326.976 meters
Distance covered by the second object:
S2 = u2 * t2 + (1/2) * a * t2^2
S2 = 10 * 1.02 + (1/2) * (-9.8) * (1.02)^2
S2 ≈ -4.9996 meters (approximately)
It is important to note that the distance covered by the second object is negative, indicating that it is below the starting point.
Finally, we can add the initial height (assuming it is the same for both objects) to the distance covered by each object to find the height above the starting point where they meet:
Height above the starting point = S1 + S2
Height above the starting point ≈ 326.976 + (-4.9996)
Height above the starting point ≈ 321.9764 meters
Therefore, the objects will meet approximately 321.9764 meters above the starting point, and they will meet after approximately 7.14 seconds from the projection of the second object.