Asked by Gordon
From a platform an object is projected vertically into air. Its height, h metres above ground level after t seconds is modelled by equation h=3+20t-5t^2.
At what height above the ground is the platform from which the object was projected?
Calculate the time when the object will again be at the same height as the platform
At what time, correct to 2 decimal places does the object reach the ground?
At what height above the ground is the platform from which the object was projected?
Calculate the time when the object will again be at the same height as the platform
At what time, correct to 2 decimal places does the object reach the ground?
Answers
Answered by
Reiny
a) at that time, t = 0, so
h = 3 + 0 + 0 = 3 m
b) you want the t, when h = 3
3 = 3 + 20t - 5t^2
5t^2 - 20t = 0
t(5t - 20) = 0
t = 0 or t = 4
c) when it hits the ground, h = 0
5t^2 - 20t - 3 = 0
solve, using your favourite method of solving quadratics, reject the negative answer.
h = 3 + 0 + 0 = 3 m
b) you want the t, when h = 3
3 = 3 + 20t - 5t^2
5t^2 - 20t = 0
t(5t - 20) = 0
t = 0 or t = 4
c) when it hits the ground, h = 0
5t^2 - 20t - 3 = 0
solve, using your favourite method of solving quadratics, reject the negative answer.
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