a particle is dropped under gravity from rest from a height and it travels a distane of 9h/25 in the last second. Calculate the height h?
Let the total fall time be t.
h = (g/2) t^2
(16/25)h = (g/2)(t-1)^2]
16/25 = [(t-1)/t]^2
Take the positive square root of both sides.
4/5 = (t-1)/t
t = 5
h = (g/2)*25 = 122.5 meters
To calculate the height, we can use the kinematic equation for uniformly accelerated motion:
s = ut + (1/2)at^2
Where:
s = distance traveled
u = initial velocity (0 m/s since the particle is dropped from rest)
a = acceleration due to gravity (-9.8 m/s^2, since it is acting in the opposite direction of motion)
t = time
From the problem, we know that in the last second (t = 1s), the particle traveled a distance of 9h/25. Let's plug these values into the equation and solve for h:
(9h/25) = (0)(1) + (1/2)(-9.8)(1)^2
(9h/25) = 0 - 4.9(1)
(9h/25) = -4.9
To solve for h, we multiply both sides of the equation by (25/9):
h = -4.9 * (25/9)
h ≈ -13.61
However, since height cannot be negative, we take the absolute value and the final height is:
h ≈ 13.61 meters
To solve this problem, we can use the equations of motion for an object under constant acceleration due to gravity.
Let's break down the problem step by step:
1. First, determine the equation of motion for the particle. In this case, the particle starts from rest, so its initial velocity (u) is 0, and it is accelerated by gravity (g) towards the ground. The equation of motion we'll use is:
s = ut + (1/2)gt^2
where s is the displacement, u is the initial velocity, t is the time, and g is the acceleration due to gravity.
2. We know that in the last second of the particle's motion, it covers a distance of 9h/25. Let's denote this distance as D. Therefore, we can substitute the values into the equation of motion:
D = u(1) + (1/2)g(1)^2
3. Since the initial velocity (u) is 0, the equation simplifies to:
D = (1/2)g
Rearrange the equation to solve for g:
g = 2D
4. We know that the time in the last second is 1 second. The particle starts from rest and falls for a period of 1 second. Therefore, in the last second, the particle covers a distance of s = (1/2)gt^2, where t = 1 second. Substitute the values:
9h/25 = (1/2)(2D)(1)^2
Simplify:
9h/25 = D
5. Now, substitute the value of D back into the equation in step 3:
g = 2D = 2(9h/25) = 18h/25
6. Finally, substitute the value of g into the original equation of motion:
s = ut + (1/2)gt^2
h = 0 + (1/2)(18h/25)(1)^2
Simplify and solve for h:
h = (1/2)(18h/25)
h = 9h/25
Multiply both sides of the equation by 25 to eliminate fractions:
25h = 9h
Subtract 9h from both sides:
25h - 9h = 9h - 9h
16h = 0
Since 16h equals zero, it implies that h must be zero.
Therefore, the height h must be zero.