An object is thrown vetically upward with an initial velocity of 14.7 m/s.How long does it take for the object to be 9.8 m from its initial position?What is the instantaneous velocity at 9.8 m from its initial position. Explain why two answers arise.
IT goes up, and comes down. That is the source of two solutions.
Hf=hi+vi*t-4.9t^2
9.8=0+14.7t-4.9t^2 solve for t.
To solve this problem, we can use the equations of motion for an object in free fall and assume that the only force acting on the object is gravity.
Step 1: Determine the time it takes for the object to reach a height of 9.8 m.
We can use the equation for the displacement of an object in free fall:
s = ut + (1/2)gt^2
where s is the displacement, u is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Rearranging the equation to solve for time, we get:
t = (sqrt(2s/g) - u/g)
Substituting the given values, u = 14.7 m/s and s = 9.8 m, we can calculate the time it takes for the object to reach a height of 9.8 m:
t = (sqrt(2*9.8/9.8) - 14.7/9.8)
t = (sqrt(2) - 1.5)
t ≈ 0.586 seconds (rounded to three decimal places)
Therefore, it takes approximately 0.586 seconds for the object to be 9.8 m from its initial position.
Step 2: Determine the instantaneous velocity at 9.8 m from its initial position.
To determine the instantaneous velocity at a specific height, we can use the equation for velocity in free fall:
v = u + gt
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
Using the given values, u = 14.7 m/s and t = 0.586 seconds, we can calculate the instantaneous velocity at 9.8 m:
v = 14.7 + 9.8*0.586
v ≈ 19.16 m/s (rounded to two decimal places)
The instantaneous velocity at 9.8 m from its initial position is approximately 19.16 m/s.
Explanation for two answers arising:
When the object is thrown upwards, it reaches its highest point and then starts falling back down. At the highest point, the object momentarily comes to a stop before changing direction. This means that at a certain height (in this case, 9.8 m), there are two instances where the object reaches that height, one while going upwards and one while coming back downwards. At each of these instances, the object has a different instantaneous velocity.
To determine the time it takes for the object to be 9.8 m from its initial position, we can use the equations of motion for a vertically thrown object.
1. The equation we will use is:
d = v₀t + (1/2)at²
where:
- d is the displacement (change in position) from the initial position,
- v₀ is the initial velocity,
- t is the time,
- a is the acceleration.
2. Given that the object is thrown vertically upward, we know that the acceleration due to gravity, a, is -9.8 m/s² (negative because it acts downward).
3. Plug in the known values into the equation:
9.8 m = 14.7 m/s * t + (1/2)(-9.8 m/s²)t²
4. Simplify the equation:
9.8 m = 14.7 m/s * t - 4.9 m/s² * t²
5. Rearrange the equation to be a quadratic equation form:
4.9 m/s² * t² - 14.7 m/s * t + 9.8 m = 0
Now, we can solve this quadratic equation using multiple methods. We can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Using the values from our equation:
a = 4.9 m/s²
b = -14.7 m/s
c = 9.8 m
Plugging these values into the formula:
t = (-(-14.7) ± √((-14.7)² - 4 * 4.9 * 9.8)) / (2 * 4.9)
t = (14.7 ± √(216.09 - 191.2)) / 9.8
t = (14.7 ± √24.89) / 9.8
t ≈ 0.2 s or 3 s
Now, let's address the second part of the question, the instantaneous velocity at 9.8 m from the initial position.
To find the instantaneous velocity, we need to determine the velocity of the object at that exact position. We can use the formula:
v = v₀ + at
Given that the initial velocity (v₀) is 14.7 m/s and the acceleration (a) is -9.8 m/s², we can calculate the velocity at 9.8 m using either of the possible times obtained earlier.
If we use t = 0.2 s:
v = 14.7 m/s + (-9.8 m/s²)(0.2 s)
= 12.7 m/s
If we use t = 3 s:
v = 14.7 m/s + (-9.8 m/s²)(3 s)
= -18.3 m/s
The two possible answers for the instantaneous velocity, 12.7 m/s and -18.3 m/s, arise because at a distance of 9.8 m from the initial position, the object can either be moving upward (positive velocity) or be at the highest point of its motion, about to start falling downward (negative velocity). This depends on whether the time calculated is during the ascending or descending motion of the object.