a pair of glasses is dropped from the top of a 36.0 m high stadium. a pen is dropped from the same position 2.21 seconds later. how high above the ground is the pend when the glasses hit the ground? (disregard air resistence. a=-g=-9.81 meters per second squared)
Calculate how long it takes the glasses to hit the gound.
36.0 = (g/2) t^2
t = 2.71 s
The pen will have been falling 2.71 - 2.21 = 0.50 seconds when the glasses hit the ground. Use that information to compute where pen is at that time.
You do that part.
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A pair of glasses are dropped from the top of a 32.0 m stadium. A pen is dropped 2.00 s later. How high above the ground is the pen when the glasses hit the ground?
Well, let's see. The glasses were dropped from a height of 36.0 m, and the pen was dropped 2.21 seconds later. So, by the time the pen starts falling, the glasses have already been falling for 2.21 seconds.
Now, I must admit, I'm not much of a mathematician. But I do know a thing or two about gravity's sense of humor. Gravity pulls objects downward at a rate of 9.81 m/s². So, after 2.21 seconds, the glasses would have fallen a distance of (9.81 * 2.21²) meters.
To find the height of the pen when the glasses hit the ground, we subtract this distance from the initial height of 36.0 m. That gives us a humorous answer. Drum roll, please...
The height of the pen when the glasses hit the ground is approximately 1.234 meters!
Just remember, gravity may bring you down, but a good sense of humor will always keep you up!
To solve this problem, we can use the kinematic equations of motion to determine the height of the pen when the glasses hit the ground.
Let's first calculate the time it takes for the glasses to reach the ground. We know that the height of the stadium is 36.0 m and the acceleration due to gravity is -9.81 m/s² (taking downward direction as negative). We can use the following equation:
h = ut + (1/2)at²
Where:
h = height (36.0 m)
u = initial velocity (0 m/s since the glasses are dropped)
t = time taken (to be determined)
a = acceleration due to gravity (-9.81 m/s²)
Substituting the given values, we get:
36.0 = 0*t + (1/2)(-9.81)t²
Rearranging the equation, we have:
-4.905t² = 36.0
Dividing both sides by -4.905, we get:
t² = -36.0 / -4.905
t² ≈ 7.34
Taking the square root of both sides, we find:
t ≈ √7.34
t ≈ 2.71 seconds (rounded to two decimal places)
Therefore, it takes approximately 2.71 seconds for the glasses to hit the ground.
Next, we need to find the time taken for the pen to hit the ground after the glasses were dropped. Given that the pen is dropped 2.21 seconds later, we can subtract this time from the time it took for the glasses to hit the ground.
Time taken for the pen to hit the ground = 2.71 - 2.21
Time taken for the pen to hit the ground ≈ 0.50 seconds
Now, we can calculate the height of the pen when it hits the ground. Again, we use the equation:
h = ut + (1/2)at²
Where:
h = height (to be determined)
u = initial velocity (0 m/s since the pen is dropped)
t = time taken (0.50 seconds)
a = acceleration due to gravity (-9.81 m/s²)
Substituting the values, we have:
h = 0*(0.50) + (1/2)(-9.81)(0.50)²
Simplifying the equation, we get:
h = 0 - 1.2275
h ≈ -1.23 m
Since the height cannot be negative, we ignore the negative sign in this case. Therefore, the height of the pen when it hits the ground is approximately 1.23 meters above the ground.