Answer:
A stone is thrown straight up from the edge of a roof, 750 feet above the ground at a speed of 14 feet per second. A) Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later? B) At what time does the stone hit the ground? C) What is the velocity of the stone when it hits the ground?
Step-by-step explanation:
A stone is thrown straight up from the edge of a roof, 750 feet above the ground at a speed of 14 feet per second. A) Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later? B) At what time does the stone hit the ground? C) What is the velocity of the stone when it hits the ground?
Answer:
37.04 m/s
Step-by-step explanation:
we can use the equation x = x0 + v0t + 1/2at^2
a = -9.8 m/s^2, the acceleration due to gravity
x0 = 70 m, the starting point of x
v0 = 0 m/s
x = 0
this will allow us to solve for time
0 = 70 + 0 +-4.9t^2
t = 3.779
now we can use the equation v = v0 + at
v = 0 + 9.8(3.77)
v = 37.04 m/s
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0.88x distributive property and isolate the variables
