A stone is dropped from the upper observation deck of a tower, 200 m above the ground. (Assume g = 9.8 m/s^2.)
a)Find the distance (in meters) of the stone above ground level at time t.
b)With what velocity does it strike the ground? (Round your answer to one decimal place.)
c)If the stone is thrown downward with a speed of 2 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)
To solve these questions, we can use the equations of motion.
a) The formula we can use to solve this question is h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity, and t is the time.
Given that the stone is dropped from a height of 200 m, we can substitute these values into the equation:
h = (1/2)gt^2
200 = (1/2)(9.8)t^2
Now, let's solve for t.
First, let's simplify the equation:
400 = 9.8t^2
Next, let's isolate t^2:
t^2 = 400/9.8
Now, take the square root of both sides of the equation:
t = √(400/9.8)
t ≈ 6.37 seconds
So, at time t, the stone is approximately 6.37 meters above the ground.
b) To find the velocity at which the stone strikes the ground, we can use the equation v = gt, where v is the final velocity and g is the acceleration due to gravity.
Using the value of g = 9.8 m/s^2 and the time it took to reach the ground, t ≈ 6.37 seconds, we can calculate the velocity:
v = gt
v = 9.8 * 6.37
v ≈ 62.69 m/s
Therefore, the stone strikes the ground with a velocity of approximately 62.69 m/s.
c) If the stone is thrown downward with an initial velocity of 2 m/s, we need to determine how long it takes for the stone to reach the ground.
We can use the same equation as before, h = (1/2)gt^2, but this time we know that h = 200 m and g = 9.8 m/s^2. We also need to find the time t, so we can rearrange the equation to solve for t.
200 = (1/2)gt^2
Rearranging the equation, we get:
t^2 = 2(200)/g
t^2 = 400/g
Now, substitute the value of g = 9.8 m/s^2:
t^2 = 400/9.8
Take the square root of both sides of the equation:
t = √(400/9.8)
t ≈ 6.37 seconds
Therefore, it takes approximately 6.37 seconds for the stone to reach the ground when thrown downward with a speed of 2 m/s.
a) distance = -4.9t^2 + 200
b) when it hits the ground, distance = 0
4.9t^2 = 200
t = √(200/4.9) = ....
d(distance)/dt = -9.8t
plug in your value of t
c) change the equation to
distance = -4.9t^2 - 2t + 200
solve using the quadratic formula