A stone is dropped from the roof of a high building. A second stoneis dropped 1.

Question

A stone is dropped from the roof of a high building. A second stoneis dropped 1.00 s later. How far apart arethe stones when the second one has reached a speed of 11.0 m/s?
Enter a number. 1 m Enter a number.

Explanation / Answer

g is the gravitational constant. acceleration a = g velocity       v = gt y = gt2/2   where y is the total distancetraveled v2(t2) = 11 m/s = g*t2 --->t2 = 11/g y2(t2) = g*(11/g)2/2 = 121/(2g) =distance traveled by the second stone when it has reached 11m/s t1 = t2 + 1 y1(t1) = g*(11/g+1)2/2 = 121/(2g)+22/2 +1*g/2 = distance traveled by the first stone when thesecond stone has reached 11 m/s. y1-y2 = 121/(2g)+11+g/2-121/(2g) =11+g/2 meter apart

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