A bal was thrown vertically upwards so that its height h metres above the ground, t seconds after it had been thrown, was given by the formula.
h=7 + 11t - 5t^2
Find
(a)the height above the ground from which the was thrown.
(b)the two values of t for which the ball was 9 metres above the ground.
h=7 + 11t - 5t^2
Find
(a)the height above the ground from which the was thrown.
(b)the two values of t for which the ball was 9 metres above the ground.
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(a) The initial height is when t = 0.
=> h(0) = 7 metres
(b) We require when h(t) = 9
=> 7 + 11t - 5t² = 9
=> 5t² - 11t + 2 = 0
i.e. (5t - 1)(t - 2) = 0
so, t = 1/5 and t = 2
The height is 9 metres at times of 0.2 seconds and 2 seconds.
:)>
=> h(0) = 7 metres
(b) We require when h(t) = 9
=> 7 + 11t - 5t² = 9
=> 5t² - 11t + 2 = 0
i.e. (5t - 1)(t - 2) = 0
so, t = 1/5 and t = 2
The height is 9 metres at times of 0.2 seconds and 2 seconds.
:)>
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(b) 2 seconds and 0.20 second
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a) 7 metres as h=7 when t=0
b) 9 = 7 + 11t - 5t^2 ie 5t^2 -11t +2 =0 ie (5t-1)(t-2)=0 ie t = 1/5sec or 2sec
b) 9 = 7 + 11t - 5t^2 ie 5t^2 -11t +2 =0 ie (5t-1)(t-2)=0 ie t = 1/5sec or 2sec