The soccer ball is kicked at 35° from the edge of the building with an initial velocity of 17 m/s and lands 71 meters away from the wall. How tall is the building that the child is standing on?
answer: m
How tall is the building that the child is standing on?
2016-09-12 7:50 am
回答 (1)
2016-09-12 8:03 am
Along horizontal direction (uniform velocity motion) :
v(x) = 17 cos35° m/s
s(x) = 71 m
s(x) = v(x) t
71 = (17 cos35°) t
t = 71 / (17 cos35°) s
Along vertical direction (uniform acceleration motion) :
(Take downward quantities as positive.)
u(x) = -17 sin35° m/s
t = 71 / (17 cos35°) s
a = g = 9.81 m/s²
s = ut + (1/2)at²
s = (-17 sin35°) × [71 / (17 cos35°)] + (1/2) × 9.81 × [71 / (17 cos35°)]²
Height of the building, s = 78 m
v(x) = 17 cos35° m/s
s(x) = 71 m
s(x) = v(x) t
71 = (17 cos35°) t
t = 71 / (17 cos35°) s
Along vertical direction (uniform acceleration motion) :
(Take downward quantities as positive.)
u(x) = -17 sin35° m/s
t = 71 / (17 cos35°) s
a = g = 9.81 m/s²
s = ut + (1/2)at²
s = (-17 sin35°) × [71 / (17 cos35°)] + (1/2) × 9.81 × [71 / (17 cos35°)]²
Height of the building, s = 78 m
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