A ball, thrown vertically upwards, from the ground, has its height h (in meters) expressed as a function of time t (in seconds), elapsed after the launch, by the law h(t) = 20t - 5t2. According to this information, determine the height at which the ball is 3 seconds after the throw and the maximum height reached by the ball.

a. 15 meters.

b. 20 meters.

Step-by-step explanation:

a. The height of the ball at 3 seconds. 20 * 3 - 5 * (3)^2 = 60 - 5 * 9 = 60 - 45 = 15.

The ball will be 15 meters high.

b. The maximum height reached by the ball.

To get that, we need to find the vertex of the parabola. We do so by doing -b/2a to find the x-coordinate of the vertex.

In this case, a = -5 and b = 20.

-20 / 2(-5) = -20 / -10 = 20 / 10 = 2.

Then, we find the y-coordinate by putting 2 where it says "t".

h(2) =  20(2) - 5(2)^2 = (40) - 5(4) = 40 - 20 = 20 meters.

Hope this helps!

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