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 on the right track! To find when the projectile hits the ground, you have correctly set the equation equal to zero by dividing both sides by -16. However, your factorization is not correct.
Let's go through the steps again:
Starting with the equation: y = -16t^2 + 32t + 240
Set y equal to zero, since we want to find when the projectile hits the ground:
0 = -16t^2 + 32t + 240
Now, you can divide both sides of the equation by -16 to simplify it further:
0 = t^2 - 2t - 15
To factorize this quadratic equation, you need to find two numbers that multiply to give -15 and add up to -2. In this case, the numbers are -5 and +3:
0 = (t - 5)(t + 3)
This means that either (t - 5) or (t + 3) must be equal to zero.
So we have two possible solutions:
t - 5 = 0 -> t = 5
t + 3 = 0 -> t = -3
Since time cannot be negative, we discard t = -3 as a valid solution. Therefore, the projectile will hit the ground at t = 5 seconds.
So, you were close! The correct time in seconds when the projectile hits the ground is t = 5.