An object is released to fall from height of 100m. At the same time another object is projected vertically upwards with velocity of 40m /s.
(i) Calculate the time taken before the objects meet
(ii) At what height do the objects meet
(i) The time taken before the objects meet can be calculated using the equation:
t = (2h)/(u + v)
where h is the height of the object, u is the initial velocity of the object falling, and v is the initial velocity of the object projected upwards.
In this case, h = 100m, u = 0m/s, and v = 40m/s.
Therefore, t = (2*100)/(0 + 40) = 5 seconds.
(ii) The objects will meet at a height of 100m, since the object falling from a height of 100m will take 5 seconds to reach the ground, and the object projected upwards will take 5 seconds to reach a height of 100m.
To find the time taken before the objects meet, we need to determine the time taken by each object to reach the meeting point.
Let's start by calculating the time taken by the object falling from a height of 100m.
Given:
Initial height, h = 100m
Acceleration due to gravity, g = 9.8 m/s^2
We can use the following kinematic equation to calculate the time taken (t1) for the object to fall:
h = ut + (1/2)gt^2, where u is the initial velocity
For the falling object:
h = 100m
u = 0m/s (since it is released)
g = 9.8 m/s^2
Substituting the values into the equation, we get:
100 = 0t1 + (1/2)(9.8)t1^2
Simplifying the equation:
4.9t1^2 = 100
Dividing both sides by 4.9:
t1^2 = 100/4.9
t1^2 ≈ 20.41
Taking the square root to find t1:
t1 ≈ √20.41
t1 ≈ 4.52 seconds (approx.)
Now, let's calculate the time taken by the object projected vertically upwards.
Given:
Initial velocity, u = 40m/s
Acceleration due to gravity, g = 9.8 m/s^2
We can use the following kinematic equation to calculate the time taken (t2) for the object to reach its maximum height:
v = u + gt, where v is the final velocity (0 in this case)
For the object projected upwards:
u = 40m/s
v = 0m/s
g = 9.8 m/s^2
Substituting the values into the equation, we get:
0 = 40 + (9.8)t2
Simplifying the equation:
9.8t2 = -40
Dividing both sides by 9.8:
t2 = -40/9.8
t2 ≈ -4.08 seconds (approx.)
Since we are interested in the time taken after the object is released, we consider the positive value. Therefore, the time taken by the object projected upwards is approximately 4.08 seconds.
Since the objects meet when the time taken by each is the same, we use the positive value:
(i) t = t1 = 4.52 seconds (approx.)
Now, let's find the height at which the objects meet.
To do this, we can calculate the distance traveled by the object projected upwards in the time t1.
Given:
t = t1 = 4.52 seconds
u = 40m/s
We can use the following kinematic equation to calculate the distance traveled (h2) by the object projected upwards:
h2 = ut - (1/2)gt^2
Substituting the values into the equation, we get:
h2 = (40)(4.52) - (1/2)(9.8)(4.52)^2
h2 ≈ 180.8 - 98.9
h2 ≈ 81.9m (approx.)
Therefore, the objects meet at a height of approximately 81.9m.