To find the answers to these questions, we will utilize the provided function h(t) = -5t^2 + 15t + 50, which represents the height of the ball above the ground at any given time t.
a) How tall is the building?
To determine the height of the building, we need to find the value of h when t is 0. Plugging t = 0 into the equation gives us:
h(0) = -5(0)^2 + 15(0) + 50
h(0) = 0 + 0 + 50
h(0) = 50
Therefore, the building's height is 50 meters.
b) How long does it take the ball to reach the ground?
The ball hits the ground when its height is h = 0. We need to find the time at which h(t) = 0. To do this, we set the equation equal to zero and solve for t:
0 = -5t^2 + 15t + 50
This is a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from our equation, we have:
t = (-(15) ± √((15)^2 - 4(-5)(50))) / (2(-5))
t = (-15 ± √(225 + 1000)) / (-10)
t = (-15 ± √(1225)) / (-10)
t = (-15 ± 35) / (-10)
The two solutions are:
t1 = (-15 + 35) / (-10) = 20 / (-10) = -2
t2 = (-15 - 35) / (-10) = -50 / (-10) = 5
Since time cannot be negative in this context, we discard the negative solution. Therefore, the ball takes 5 seconds to reach the ground.
c) For how long is the ball 55m above the ground?
We need to find the time (t) when the height of the ball (h) is 55 meters. We set h(t) = 55 and solve for t:
55 = -5t^2 + 15t + 50
Re-arranging the equation to form a quadratic equation:
5t^2 - 15t + 5 = 0
Solving this equation using factoring, completing the square, or the quadratic formula gives us two solutions. Let's use the quadratic formula again:
t = (-(-15) ± √((-15)^2 - 4(5)(5))) / (2(5))
t = (15 ± √(225 - 100)) / 10
t = (15 ± √(125)) / 10
t = (15 ± √(5^2 * 5)) / 10
t = (15 ± 5√5) / 10
The two solutions are:
t1 = (15 + 5√5) / 10
t2 = (15 - 5√5) / 10
Therefore, the ball is 55 meters above the ground for t1 seconds and t2 seconds, where t1 and t2 are the solutions obtained above.