When thrown into the air from the top of a 50 ft building, a ball’s height, S, at time t can be found by S(t) = -16t^2 + 32t + 50. When t = 1, s = -16(1)^2 + 32(1) + 50 = 66. This implies that after 1 second, the height of the ball is 66 feet. When t = 2, s = -16(2)^2 + 32(2) + 50 = 50. This implies that after 2 seconds, the height of the ball is 50 feet. could someone break this down so I can understand it.

thats an example of your assignment, your suppose to come up with your on real life application. I'm in this class also.

Sure! Let's break down the problem step by step:

1. The equation S(t) = -16t^2 + 32t + 50 represents the height of the ball at time t. In this equation, t represents the time in seconds and S(t) represents the height in feet.

2. In the first part, when t = 1, we need to find the height of the ball after 1 second. To do this, we substitute t = 1 into the equation: S(1) = -16(1)^2 + 32(1) + 50. Simplifying this gives us S(1) = -16 + 32 + 50 = 66. This means that after 1 second, the height of the ball is 66 feet.

3. Similarly, in the second part, when t = 2, we need to find the height of the ball after 2 seconds. We substitute t = 2 into the equation: S(2) = -16(2)^2 + 32(2) + 50. Simplifying this gives us S(2) = -64 + 64 + 50 = 50. This means that after 2 seconds, the height of the ball is 50 feet.

So, this is how we calculate the height of the ball at different times using the given equation. We substitute the value of t into the equation and solve for S(t) to find the height in feet.