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Take the vertical upward direction as the positive direction, when the stone moves to the top of the throwing point and is 15m away from the throwing point, the displacement is x=15m, which is changed by x=v0t-12gt
Substituting yields 15=20t-1
2 10t solution yields t1=1s, t2=3s
When the stone moves to 15m below the throwing point and is 15m away from the throwing point, the displacement is x=-15m, which is changed by x=v0t-12gt
Substituting yields 15=20t-1
2 10t solution gives t1=(2+
7)s,t2=(2-
7) s (to be discarded).
Therefore, ACD is chosen
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15=20t-1
Á10t¬½t1
1s£¬t2
15=20t-1
Á10t¬½t1
No air resistance is assumed:
According to v=gt, the time is: 4 seconds for the round trip; >>>More
The first case: 15m above the throw point.
The equation is: 20t-5t 2=15 >>>More
From the starting point down to 15m is the time SQRT(7)-2; Coupled with the above round-trip time, 4s is a total of 2+sqrt(7) seconds.
25=30t-1/2×10t²
25=30t-5t² >>>More
Ascending Stage:
According to the formula: V=V0-gt, and the formula H= >>>More