An object is dropped from the top of a tall building. at 2 seconds, it is 64 feet from the top of the building. at 4 seconds, it is 256 feet from the top of the building. what is the average rate the object was traveling in the interval between 2 and 4 seconds?
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Ответ:
96.21 ft/s
Explanation:
To solve this, you only need to use one expression which is:
Vf² = Vo² + 2gh
g = 9.8 m/s²
However, this exercise is talking in feet, so convert the gravity to feet first:
g = 9.8 * 3.28 = 32.15 ft/s²
Vo is zero, because it's a free fall and in free fall the innitial speed is always zero. With this, let's calculate the speed at 2 seconds, with a height of 64 ft, and then with the 256 ft:
V1 = √2*32.15*64
V1 = 64.15 ft/s
V2 = √2*32.15*256
V2 = 128.3 ft/s
So the average rate is:
V = 128.3 + 64.15 / 2
V = 96.22 ft/s
Ответ:
0
Step-by-step explanation:
y=10+z=5
z=5-10
z=-5
y=z+5
Substitute for y
0.3 y + y / Z
0.3(z+5)+(z+5)/z
z=-5
0.3(-5+5)+(-5+5)/-5
0.3(0)+0/-5
0.3/0
0