Based on the equation y=-5x^2+10x, we can see that this is a quadratic equation in the form of a downward facing parabola. This indicates that the ball was kicked from the ground, as the initial height (y-intercept) is 0.
To find the horizontal distance at which the ball hits the ground, we set y=0 and solve for x:
0 = -5x^2 + 10x
0 = x(-5x + 10)
0 = x(10 - 5x)
x=0 or x=2
So, the ball hit the ground at a horizontal distance of 2 metres.
To find the maximum height of the ball, we can use the formula for the x-coordinate of the vertex of a parabola: x = -b/(2a). In this case, a=-5 and b=10.
x = -10 / (2*-5) = -10 / -10 = 1
Now we substitute x=1 back into the original equation to find the maximum height:
y = -5(1)^2 + 10(1) = -5 + 10 = 5
Therefore, the maximum height of the ball is 5 metres.
A soccer ball was kicked during a game. The path of the soccer ball can be defined by the relation y=-5x^2+10x, where y represents the height in metres and x represents the horizontal distance in metres travelled by the ball. Do you think the ball was kicked from in the air? At what horizontal distance did the ball travel before hit the ground? What was the maximum height of the ball?
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