A body dropped from the top of the tower covers a distance 7x in the last second of its journey, where x is the distance covered in first second. How much time does it take to reach the ground?
Let's assume the time taken to reach the ground is "t" seconds.
In the first second, the distance covered is "x".
In the last second, the distance covered is 7x.
From this information, we can set up an equation:
Distance covered in the first second + Distance covered in the last second = Total distance
x + 7x = Total distance
Simplifying the equation, we have:
8x = Total distance
Now, we know that the total distance covered is equal to the height of the tower. Let's assume the height of the tower is "h".
Therefore,
8x = h
Now, we need to find the time taken to reach the ground, which is "t" seconds.
Using the formula for distance traveled with constant acceleration:
Distance = (Initial velocity × Time) + (0.5 × Acceleration × Time^2)
Since the body is only falling under the influence of gravity, the initial velocity is 0 and the acceleration is due to gravity, which is approximately 9.8 m/s^2.
So, the distance traveled in time "t" is:
h = (0 × t) + (0.5 × 9.8 × t^2)
h = 0 + 4.9t^2
Now we can substitute "h" with "8x" in this equation:
8x = 4.9t^2
Simplifying the equation:
t^2 = (8x) / 4.9
t^2 = 1.632653 x
Taking the square root of both sides:
t = sqrt(1.632653 x)
Therefore, it takes approximately sqrt(1.632653 x) seconds for the body to reach the ground.
To determine the time it takes for the body to reach the ground, we can make use of equations of motion. Let's break down the problem step by step:
1. Let's assume that the distance covered by the body in the first second is x.
2. According to the problem, the distance covered by the body in the last second is 7x.
3. We need to find the total time it takes for the body to reach the ground.
Now, let's use the equation of motion to find the time taken for the body to fall from the top of the tower to the ground:
The equation of motion for distance covered by a falling body is given by:
s = ut + (1/2)at^2
Where:
- s is the distance covered
- u is the initial velocity (which is zero in this case since the body is just dropped)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken
Using this equation, we can first find the time it takes for the body to cover the distance x in the first second:
x = 0*t + (1/2)*(-9.8)*t^2
x = (1/2)*(-9.8)*t^2
Similarly, for the last second:
7x = 0*t + (1/2)*(-9.8)*t^2
7x = (1/2)*(-9.8)*t^2
We have two equations, one for the first second and one for the last second. Since we know that the distance covered in the last second is 7 times the distance covered in the first second, we can set up an equation:
7x = (1/2)*(-9.8)*t^2
Dividing both sides by x:
7 = (-4.9)*t^2
Dividing both sides by -4.9:
1.42857 = t^2
Taking the square root of both sides:
t = 1.195 seconds (rounded to 3 decimal places)
Therefore, it takes approximately 1.195 seconds for the body to reach the ground.
Average velocity in first second = g/2 = 4.9 m/s
Average velocity in 2nd second = 3g/2 = 14.7 m/s
Average velocity in 3rd second = 5g/2 = 24.5 m/s
Average velocity in 4th second = 7g/2 = 34.2 m/s, which is seven times the first second.
It takes 4 seconds.