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27.07.2019 •
Mathematics
Path of a ball after kicked from a height of .5 m above ground is given by h=-0.1d^2+1.1d+.5 (h-height above ground, d-horizontal distance)? how far has the ball travelled horizontally, to the nearest tenth metre, when it lands?
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Ответ:
h = -0.1 d^2 + 1.1d + 0.5
When the ball lands, its height becomes zero, therefore:
To get the distance, we will simply substitute with h=0 in the equation and then factorize it to get the possible distance values as follows:
0 = -0.1 d^2 + 1.1d + 0.5
(x-11.44)(x+0.44) = 0
either the distance x = 11.44 m (accepted value)
or x = -0.44 m (rejected value as distance cannot be 0)
Ответ:
.
This is a function of the height of the ball, in terms of d, the horizontal distance.
When the ball lands, h is equal to 0, so we need to find the value of d for which h(d) is 0, so we need to solve the equation:
.
Thus, we have a quadratic equation to solve: we use the discriminant formula!
a=-0.1, b=1.1, c=0.5, thus the discriminant is .
The square root of 1.41 is approximately 1.19,
thus, the roots of the equation are:
stance must be positive, so we only consider the second answer. Thus, d=11.5 m
11.5
Ответ: