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shjblover812
11.12.2019 •
Mathematics
During a baseball game the batter hits a high pop-up directly over home plate. if the ball stays in the air sixty seconds, how high did the ball go
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Ответ:
h = height (m) at time t (s).
Because the ball stays in the air for 60 s, the ball reaches a maximum height after 30 s, when the upward velocity is zero.
Note that
g = 9.8 m/s², the acceleration due to gravity.
Let v = the vertical launch velocity. Then
v - (9.8 m/s²)*(30 s) = 0
v = 294 m/s
Also, the maximum height, H, is
H = (294 m/s)*(30 s) - (1/2)*(9.8 m/s²)*(30 s)²
= 4410 m
4410 m
Ответ:
x-2y=-7
as they x and 2y are variables they be in one side living the one with no variable