The expression - 5t^2 + 40t predicts the height, in meters, of an object t seconds after a person launches it into the air. Using this expression, how many seconds will it take the object to hit the ground?
How do I solve this?
when it hits the ground, the height is zero, right?
So, just solve
-5t^2 + 40t = 0
-5t(t-8) = 0
...
Well, if you want to figure out when the object hits the ground, you'll need to find the value of t when the height is equal to zero. In other words, you're looking for the root of the expression -5t^2 + 40t = 0.
Now, let me tell you a little joke to ease the math anxiety. Why don't scientists trust atoms? Because they make up everything!
Okay, back to the problem. To solve this equation, you can use factoring, the quadratic formula, or completing the square. But since I'm in a funny mood, I'm going to go with the quadratic formula.
So, the quadratic formula says that for an equation of the form ax^2 + bx + c = 0, the solution for x is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -5, b = 40, and c = 0. Plug those values into the quadratic formula, simplify, and calculate the roots. The positive root will give you the time it takes for the object to hit the ground.
But remember, while you're crunching those numbers, hold on to your sides with laughter, so you won't fall down before the object does!
To find the time it takes for the object to hit the ground, we need to determine when the height is equal to zero. We can do this by solving the equation:
-5t^2 + 40t = 0
Step 1: Factor out a common factor, -5t:
(-5t)(t - 8) = 0
Step 2: Set each factor equal to zero and solve for t:
-5t = 0 --> t = 0
t - 8 = 0 --> t = 8
The solution t = 0 corresponds to when the object was launched, so we disregard it. Therefore, the object will hit the ground after 8 seconds.
To solve this problem, you need to find the value of t when the height is 0, which represents the moment the object hits the ground.
We can set the expression for height, -5t^2 + 40t, equal to 0 and solve for t. The equation becomes:
-5t^2 + 40t = 0
Next, we can factor out common terms to simplify the equation:
t(-5t + 40) = 0
Now, we have two factors to consider: t = 0 and -5t + 40 = 0.
Solving the first factor t = 0, we can see that this represents the initial launch of the object. Therefore, we ignore this solution for our case since we are interested in when the object hits the ground after being launched.
To solve the second factor, -5t + 40 = 0, we can isolate t by moving 40 to the other side of the equation:
-5t = -40
Divide both sides of the equation by -5 to solve for t:
t = -40 / -5 = 8
Therefore, the object will hit the ground after approximately 8 seconds.