A projectile is launched vertically upward from the top of a 240-foot building. This translates 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 the projectile hit the ground?
Duplicate post. See my previous answer.
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To find the time when the projectile hits the ground, we need to set the height (y) to zero and solve for t in the equation:
y = -16t^2 + 32t + 240
Setting y = 0:
0 = -16t^2 + 32t + 240
Now we can solve for t by factoring or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 32, and c = 240. Substituting these values into the formula:
t = (-32 ± √(32^2 - 4(-16)(240))) / (2(-16))
Simplifying inside the square root:
t = (-32 ± √(1024 + 15360)) / (-32)
t = (-32 ± √(16384)) / (-32)
t = (-32 ± 128) / (-32)
Now we have two possible solutions:
1. t = (-32 + 128) / (-32)
t = 96 / -32
t = -3
2. t = (-32 - 128) / (-32)
t = -160 / -32
t = 5
Since time cannot be negative in this case, we discard the first solution (-3). Therefore, the projectile will hit the ground after 5 seconds.
To determine when the projectile will hit the ground, we need to find the time at which the height y becomes zero.
Given the equation y = -16t^2 + 32t + 240, we can set y equal to zero:
0 = -16t^2 + 32t + 240
Now we have a quadratic equation. We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 32, and c = 240. Plugging these values into the quadratic formula, we get:
t = (-32 ± √(32^2 - 4(-16)(240))) / (2(-16))
Simplifying this expression further, we have:
t = (-32 ± √(1024 + 15360)) / -32
t = (-32 ± √(16384)) / -32
t = (-32 ± 128) / -32
Now we have two possible solutions for t:
t1 = (-32 + 128) / -32 = 96 / -32 = -3
t2 = (-32 - 128) / -32 = -160 / -32 = 5
Since time cannot be negative in this context, we ignore the negative solution.
Therefore, the projectile will hit the ground after 5 seconds.