A ball is kicked into the air and follows a path described by h(t)=-4.9t^2 +8.4t+0.6, where t is the time in seconds and h is the height of the balls in metres. Determine the maximum height of the ball, to the nearest tenth of a metre.
This is a parabola, and the peak will be halfway between the roots.
so the roots are...
0=4.9t^2 -8.4t-0.6
0 =t^2 -1.71t-0.122
t=(1.71/2 +- sqrt....
so the midpoint, or the max height occures at t=1.71/2=.86 seconds.
then height is h(t)=-4.9t^2 +8.4t+0.6 with t=above.
If you are in calculus, it is much easier, find when it stop rising upward.
dh/dt=0=-9.8t+8.4, solve for t..then solve for h(that t).
1.71+-sqrt(2.92-.488)/2
t= -.86 +-.78
halfway between these roots are -.86, so the beak occurs
To determine the maximum height of the ball, we need to find the vertex of the quadratic function h(t) = -4.9t^2 + 8.4t + 0.6.
The vertex of a quadratic function in the form h(t) = at^2 + bt + c can be found using the formula t = -b / (2a).
Here, a = -4.9 and b = 8.4.
t = -8.4 / (2*(-4.9))
t = -8.4 / (-9.8)
t ≈ 0.8571
Now, substitute this value of t back into the original equation to find the maximum height:
h(t) = -4.9(0.8571)^2 + 8.4(0.8571) + 0.6
h(t) ≈ -4.9(0.7347) + 7.199 + 0.6
h(t) ≈ -3.599 + 7.199 + 0.6
h(t) ≈ 4.2
Therefore, the maximum height of the ball is approximately 4.2 meters.
To determine the maximum height of the ball, we need to find the vertex of the parabolic function that represents the height of the ball.
The vertex of a parabola in the form h(t) = at^2 + bt + c can be found using the formula:
t = -b / (2a)
In this case, a = -4.9 and b = 8.4. Plugging these values into the formula, we get:
t = -8.4 / (2 * -4.9)
t = -8.4 / -9.8
t ≈ 0.8571
Now that we have the time at which the vertex occurs, we can substitute this value into the equation to find the maximum height.
h(t) = -4.9t^2 + 8.4t + 0.6
h(0.8571) = -4.9(0.8571)^2 + 8.4(0.8571) + 0.6
h(0.8571) ≈ 4.5
Therefore, the maximum height of the ball is approximately 4.5 meters.