Asked by Anonymous
(b) A stone is thrust upwards and reaches a height of s= (73.5t-4.9t2) metres at the
end of t seconds. Determine the highest point of the stone’s ascent. [6]
end of t seconds. Determine the highest point of the stone’s ascent. [6]
Answers
Answered by
drwls
That would be when
ds/dt = 73.5 - 9.8t = 0
t = 7.53 seconds
s = 73.5*7.53 - 4.9*(7.53)^2
= 275.6 meters
Here is another method:
(Intial kinetic energy)/m = (maximum optential enegy)/m
Vo^2/2 = gHmax
Hmax = (1/2)(73.5)^2/9.8 = 275.6 m
ds/dt = 73.5 - 9.8t = 0
t = 7.53 seconds
s = 73.5*7.53 - 4.9*(7.53)^2
= 275.6 meters
Here is another method:
(Intial kinetic energy)/m = (maximum optential enegy)/m
Vo^2/2 = gHmax
Hmax = (1/2)(73.5)^2/9.8 = 275.6 m
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