A body thrown vertically up from the ground passes the height 10.2m twice at an interval of 10 sec. Its initial velocity was (g = 10 m/s2)
To solve this problem, we can use the equations of motion for vertical motion with constant acceleration. Let's break it down step by step:
Step 1: Identify the knowns and unknowns:
Knowns:
- The acceleration due to gravity, g = 10 m/s^2
- The change in height, Δh = 10.2 m
- The time interval, Δt = 10 s
Unknown:
- The initial velocity, u
Step 2: Use the kinematic equation to calculate the initial velocity:
The equation we can use is:
Δh = ut + (1/2)gt^2
Since the body passes the height twice, we can consider the time taken for each ascent separately:
For the first pass, Δh1 = 10.2 m and t1 = 10 s
For the second pass, Δh2 = 10.2 m and t2 = 20 s
Equation for the first pass:
10.2 = (u * 10) + (1/2)(10)(10^2)
10.2 = 10u + 500
10u = -489.8
Equation for the second pass:
10.2 = (u * 20) + (1/2)(10)(20^2)
10.2 = 20u + 2000
20u = -1989.8
Step 3: Solve the equations for the initial velocity:
From the first equation:
10u = -489.8
u = -48.98 m/s
From the second equation:
20u = -1989.8
u = -99.49 m/s
Note: We have two different values for the initial velocity; one is positive, and the other is negative. This indicates that the body was either thrown up or down. However, we assumed the body was thrown vertically up, so the only possible solution is u = -48.98 m/s.
Therefore, the initial velocity of the body was approximately -48.98 m/s when thrown upwards.
To find the initial velocity of the body, we need to use the equation of motion that relates the height, time, and initial velocity of an object in free fall.
First, let's label the given information:
- Height achieved: 10.2 m (twice)
- Time interval: 10 seconds
We know that when the body is thrown vertically up, it will reach the same height twice during its motion. The time it takes to reach each height is given as a 10-second interval.
Now, let's consider the equation of motion for an object in free fall:
h = (v_initial * t) + (1/2 * g * t^2)
Where:
h is the height reached,
v_initial is the initial velocity of the object,
t is the time, and
g is the acceleration due to gravity (approximately 10 m/s^2).
Since the body passes the height of 10.2 m twice, we can set up the following equations:
10.2 = (v_initial * 10) + (1/2 * 10 * 10^2) -- (1)
10.2 = (v_initial * 20) + (1/2 * 10 * 20^2) -- (2)
Now, let's solve these equations to find the value of v_initial.
From equation (1):
10.2 = (10v_initial) + (50)
Simplifying equation (1):
10v_initial = 10.2 - 50
10v_initial = -39.8
v_initial = -3.98 m/s
Now, let's use equation (2) to confirm the value of v_initial:
10.2 = (20 * v_initial) + (200)
Simplifying equation (2):
20v_initial = 10.2 - 200
20v_initial = -189.8
v_initial = -9.49 m/s
Since velocity cannot be negative for an object thrown vertically upwards, we need to consider the positive value of v_initial, which is:
v_initial ≈ 9.49 m/s
Therefore, the initial velocity of the body is approximately 9.49 m/s.
mein gawar hu mujhe nahi ata
In this problem, it takes 5 seconds for the body to rise from 10.2 m elevation to the maximum height, and another 5 seconds to fall back to 10.2 m elevation.
The speed at 10.2 m height is then
V(10.2m) = g* 5 s = 50 m/s
The initial velocity Vo is whatever allows the body to have a speed of 50 m/s when the elevation is 10.2 m
Use conservation of energy.
M g *10.2 m = (M/2)[Vo^2 - 50^2]
M cancels out. Solve for Vo.