A body thrown upward from ground covers equal distance in 4th and 7th second . With what initial velocity body was projected ?
Distance in 4th second= Distance in 7th second.
Distance in 4th second= u+{a/2(2n-1)}...where n=4..
Solving this you will get
S=u-35...where a=-g=-10.(because body is moving upwards)
Same for n=7
You will get
S=u-65
Now distance will be same only when one at one time it is moving upwards and other time it is moving downwards..
So S4=-S7
U-35=-(U-65)
After solving you will get U=50m/s
To solve this problem, we can use the equations of motion related to projectile motion. Let's break down the given information and solve step by step:
Given:
Time taken to cover equal distances in the 4th and 7th second is equal.
Let's assume:
Initial velocity (u) = ?
Time taken to cover equal distances = 3 seconds (7th second - 4th second)
Acceleration due to gravity (g) = -9.8 m/s² (taking negative sign as the body is thrown upwards)
Now, we'll use the equation:
s = ut + 0.5 * a * t²
From the given information, we can calculate the distance covered by the body in the 4th and 7th second.
For the 4th second:
t = 4 seconds
s1 = ut + 0.5 * a * t²
For the 7th second:
t = 7 seconds
s2 = ut + 0.5 * a * t²
We know that the distances covered in both the 4th and 7th second are equal. So, we can equate the two equations:
s1 = s2
ut + 0.5 * a * (4)² = ut + 0.5 * a * (7)²
Simplifying further:
4u + 0.5 * (-9.8) * 16 = 7u - 0.5 * 9.8 * 49
4u - 78.4 = 7u - 240.1
Moving similar terms to one side:
7u - 4u = 240.1 - 78.4
3u = 161.7
Dividing both sides by 3:
u = 161.7 / 3
Therefore, the initial velocity of the body when it was projected is approximately 53.9 m/s.
To find the initial velocity with which the body was projected, we can use the equations of motion for vertical motion under constant acceleration.
Let's break down the problem to gather the given information:
1. The body is thrown upward from the ground.
2. The total time when it covers equal distances in the 4th and 7th second.
We know that the distance covered by a body in upward motion is given by the equation:
s = ut + (1/2)at^2
where:
s = distance covered
u = initial velocity
t = time taken
a = acceleration due to gravity (approximated as -9.8 m/s^2, taking the upward direction as positive)
First, let's find the distance covered in the 4th and 7th second.
Distance covered in the 4th second:
s_4 = u * 4 + (1/2) * (-9.8) * (4^2)
Distance covered in the 7th second:
s_7 = u * 7 + (1/2) * (-9.8) * (7^2)
Given that the distances covered in the 4th and 7th second are equal, we can form an equation:
s_4 = s_7
By substituting the respective values, we can simplify the equation:
u * 4 + (1/2) * (-9.8) * (4^2) = u * 7 + (1/2) * (-9.8) * (7^2)
Now, let's solve the equation to find the initial velocity (u).
Steps to solve the equation:
1. Expand and simplify both sides.
4u - 1/2 * 9.8 * 16 = 7u - 1/2 * 9.8 * 49
2. Group the u terms on one side and the constant terms on the other side.
7u - 4u = -(1/2) * 9.8 * 49 + (1/2) * 9.8 * 16
3. Simplify both sides.
3u = -24.5 * 49 + 4 * 9.8 * 16
4. Evaluate the right side of the equation.
3u = -1199.5 + 627.2
5. Combine the terms on the right side.
3u = -572.3
6. Divide both sides by 3 to isolate u.
u = -190.7667
Since velocity is a vector quantity, the negative sign indicates that the initial velocity was in the downward direction. However, in this case, we are interested in the magnitude of the initial velocity, so we take the absolute value:
|u| ≈ 190.77 m/s
Therefore, the approximate initial velocity with which the body was projected is 190.77 m/s.
h=vi*t-1/2 g t^2
distance fourth second=vi*5-4.8*25-vi*4+4.8*16=vi-4.8(9)
distance seventh second= similar math=vi-4.8(64-49)=vi-4.8(15)
now, distances are "equal" if one ignores direction, but considering direction, then
vi-4.8*9=-(vi-4.8*15)
2vi=4.8(6)
vi= 2.4*6=14.4
check: distance in fourth second
h(5)-h(4)=14.4*(5-4)+4.8(16-25)
= 14.4+4.8*9=57.6
distance in seventh second
h(7)-h(8)=14.4(-1)+4.8(64-49)
= -14.4+4.8(15)=57.6