A stone is dropped under gravity from rest from a height h and it travels a distance 9h/25 in the last second. What is the height H?
H=gt²/2
16H/25 =g(t-1)²/2
16gt²/2•25 = g(t²-2t+1)/2
16t²=25t²- 50t +25
9 t²-50t +25=0
t= (50±sqrt(2500-4•9•25))/18=(50±40)/18
t₁=5 s, t₂=0.55s ( extraneous root since is less than 1 second)
H= gt²/2 =9.8•5²/2=122.5 m
To solve this problem, we can use the kinematic equation for motion under constant acceleration.
The equation we need to use is:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (which is 0 since the stone is dropped from rest)
a = acceleration (which is the acceleration due to gravity, approximately -9.8 m/s^2)
t = time
In the last second of the stone's fall, the time would be 1 second. Therefore, we can rewrite the equation as:
9h/25 = 0 + (1/2)(-9.8)(1)^2
Simplifying the equation, we get:
9h/25 = -4.9
To find the value of h, we can multiply both sides of the equation by 25/9:
h = (-4.9)(25/9)
Simplifying further, we get:
h = -12.8
The negative sign indicates that the height h is measured downwards. To find the positive height, we can take the absolute value:
H = |h| = |-12.8| = 12.8
Therefore, the height H is 12.8 units.
To find the height H, we need to understand the relationship between distance and time traveled for an object under the influence of gravity.
First, we can analyze the distance traveled in the last second. The formula to calculate the distance traveled under constant acceleration is:
d = ut + (1/2)at^2
Where:
- d is the distance traveled
- u is the initial velocity (in this case, 0 m/s, as the stone is dropped from rest)
- a is the acceleration (gravity, which is approximately 9.8 m/s^2)
- t is the time traveled (1 second, as mentioned)
Substituting the given values, we get:
9h/25 = 0 + (1/2)(9.8)(1)^2
9h/25 = 4.9
Now, we can solve for h:
9h = 4.9 * 25
h = (4.9 * 25) / 9
h ≈ 13.611 m
Therefore, the height H is approximately 13.611 meters.