A projectile is launched vertically upward from the top of a 240-foot building. This translate to the model
y = -16t^2 + 32t + 240 where y is the height in feet and t is the time in seconds after it was launched. When will he projectile hit the ground?
would i divide both sides by -16?
if so
t^2-2t-15=0
which i think factors to
(t-15)(t+13)=0
t=15 seconds when it hit the ground.
I am close?
Yes, you are close, but there seems to be a small error in your factoring. Let's go through the steps again to find the correct solution:
Starting with the equation,
-16t^2 + 32t + 240 = 0
Step 1: Divide both sides by -16 to simplify the equation:
t^2 - 2t - 15 = 0
Step 2: Factor the quadratic equation:
(t - 5)(t + 3) = 0
Step 3: Set each factor equal to zero and solve for t:
t - 5 = 0 --> t = 5 seconds
t + 3 = 0 --> t = -3 seconds
Since time cannot be negative, we discard the negative solution t = -3.
Therefore, the projectile will hit the ground after approximately 5 seconds.
Yes, you're very close! However, there seems to be a small error in the equation you've factored. Let me help you correct it.
To find the time when the projectile hits the ground, we need to set y = 0 in the equation y = -16t^2 + 32t + 240, since the height of the projectile at the ground is zero.
So, the equation becomes:
0 = -16t^2 + 32t + 240
To solve this quadratic equation, you can indeed divide both sides by -16 to simplify it. Doing so gives you:
t^2 - 2t - 15 = 0
Now, let's factor this quadratic equation correctly:
(t - 5)(t + 3) = 0
Now, set each factor equal to zero:
t - 5 = 0 or t + 3 = 0
Solving these equations, we get:
t = 5 or t = -3
Since time cannot be negative in this context, we discard the negative value, t = -3.
Therefore, the projectile will hit the ground when t = 5 seconds.
Great job on your calculations, and you were indeed very close!