A ball is thrown from the ground with a
velocity of 80 ft/sec. Then the ball will be at a height of 96 feet above the ground after time..?
Answers
Answered by
19
Answer:
s = ut + 1/2 (at^2)
a = -9.8 m/s = -32.18 ft/sec
s = 96 ft
u = 80 ft/sec
So 96 = 80t + 1/2(-32.18).t^2
So we get a quadratic :
16.09t^2 - 80t + 96 = 0
(t-2.94)(t-32.65) = 0
But t can't be equal to 32.65 as the ball we reach top by 2.94 sec.
So t = 2.94 sec
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harshhjangid:
bro the ans is "2 and 3 sec".
Bro 2.94 sec can be estimated to 3 sec so the answer is 3.
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thts another option.. which is not correct.. there should me two ans. as the ball is going upwards.. it will com downward too..
but you have only asked time to reach height t
Answered by
11
Answer:2 and 3
Explanation:
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