my teacher gave me this question for review and i don't know how to do it!! HELP!!!
A ball is thrown upwards, from a building. It's height(h), in meters above the ground seconds(t) after being thrown is given by h(t)=-5t^2+15t+50
a) How tall is the building? (explain your answer)
b) How long does it take the ball to reach the ground?
c) For how long is the ball 55m above the ground?
surely you have similar question that you can study.
You know that the initial height is the constant term. So, the building is 50m tall
The ball hits the ground when the height is zero. So, just solve for t when h=0
I assume you want to know how long the ball is at least 55m above the ground. It is only at 50m for an instant. So, we want
-5t^2+15t+50 >= 55
1/2 (3-√5) <= t <= 1/2 (3+√5)
http://www.wolframalpha.com/input/?i=+-5t^2%2B15t%2B50+%3E%3D+55
I can help you with these questions. Let's break them down one by one:
a) To find how tall the building is, we need to determine the height of the ball at t=0 when it was thrown. In the equation h(t) = -5t^2 + 15t + 50, substitute t=0:
h(0) = -5(0)^2 + 15(0) + 50
= 0 + 0 + 50
= 50
Therefore, the building is 50 meters tall.
b) To find how long it takes the ball to reach the ground, we need to determine when the height of the ball is equal to zero. In other words, we need to solve the equation:
-5t^2 + 15t + 50 = 0
This is a quadratic equation. We can either factor it or use the quadratic formula to find the solutions. Using the quadratic formula, we have:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a=-5, b=15, and c=50. Substituting these values into the quadratic formula, we get:
t = (-15 ± sqrt(15^2 - 4(-5)(50))) / (2(-5))
= (-15 ± sqrt(225 + 1000)) / (-10)
= (-15 ± sqrt(1225)) / (-10)
= (-15 ± 35) / (-10)
We have two solutions: t = (-15 + 35) / (-10) = 2 and t = (-15 - 35) / (-10) = 5. The ball takes 2 seconds to reach the ground.
c) To find for how long the ball is 55m above the ground, we need to solve the equation:
-5t^2 + 15t + 50 = 55
Rearranging the equation, we get:
-5t^2 + 15t - 5 = 0
Again, we can use the quadratic formula to find the solutions. Substituting a=-5, b=15, and c=-5 into the quadratic formula, we get:
t = (-15 ± sqrt(15^2 - 4(-5)(-5))) / (2(-5))
= (-15 ± sqrt(225 + 100)) / (-10)
= (-15 ± sqrt(325)) / (-10)
However, since we are looking for a positive time value, the negative solution is not relevant. Therefore, we take only the positive solution:
t = (-15 + sqrt(325)) / (-10)
Simplifying this expression will give us the exact value for how long the ball is 55 meters above the ground.