(h=-16t^2+64t+80), after how many seconds will the object hit the ground at the bottom of the cliff.
You must solve equation:
- 16 t ^ 2 + 64 t + 80 = 0
- 16 t ^ 2 + 64 t + 80 = 0 Divide both sides by - 16
- 16 t ^ 2 / - 16 + 64 t / - 16 + 80 / - 16 = 0
t ^ 2 - 4 t - 5 = 0 Add 5 to both sides
t ^ 2 - 4 t - 5 + 5 = 0 + 5
t ^ 2 - 4 t = 5 Add 4 to both sides
t ^ 2 - 4 t + 4 = 5 + 4
t ^ 2 - 4 t + 4 = 9
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Remark :
t ^ 2 - 4 t + 4 = ( t - 2 ) ^ 2
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( t - 2 ) ^ 2 = 9 Take the square root of both sides
t - 2 = - 3
and
t - 2 = 3
t - 2 = - 3 Add 2 to both sides
t - 2 + 2 = - 3 + 2
t = - 1 s
t - 2 = 3 Add 2 to both sides
t - 2 + 2 = 3 + 2
t = 5 s
Solutions:
t = - 1 a and t = 5 s
Time cant be negative so : t = 5 s
Proof:
- 16 t ^ 2 + 64 t + 80 =
- 16 * 5 ^ 2 + 64 * 5 + 80 =
- 400 + 320 + 80 = 0
Well, it seems like the object is taking its sweet time to hit the ground! But don't worry, I'll calculate it for you. Let me put on my mathematician nose and get to work.
To find out when the object hits the ground, we need to solve the equation h = 0. So let me put my equation-solving skills into action.
(-16t^2 + 64t + 80) = 0
If we plug this into the quadratic formula (I know, it sounds fancy), we'll get a couple of solutions. So to find the time when the object hits the ground, we need to find the positive solution.
After carrying out the calculations (drumroll, please), the object will hit the ground at the bottom of the cliff after approximately 2.5 seconds. Woohoo! But remember, this is just math - in real life, gravity tends to rush things a bit faster.
To find the time when the object hits the ground, we need to find the value of t when h equals 0.
Given the equation: h = -16t^2 + 64t + 80
Setting h = 0, we have:
0 = -16t^2 + 64t + 80
To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 64, and c = 80.
Plugging these values into the quadratic formula:
t = (-64 ± √(64^2 - 4(-16)(80))) / (2(-16))
Simplifying further:
t = (-64 ± √(4096 + 5120)) / (-32)
t = (-64 ± √9216) / (-32)
t = (-64 ± 96) / (-32)
Now, we have two possible solutions:
t1 = (-64 + 96) / (-32)
t2 = (-64 - 96) / (-32)
Calculating these values:
t1 = 32 / (-32) = -1
t2 = -160 / (-32) = 5
Therefore, the object will hit the ground at the bottom of the cliff after 5 seconds.
To find out when the object hits the ground, we need to determine the value of t when the height (h) is equal to zero.
The given equation is h = -16t^2 + 64t + 80.
When the object hits the ground, it means its height is zero, so we can set h to zero and solve for t:
0 = -16t^2 + 64t + 80.
This equation is a quadratic equation in the form of ax^2 + bx + c = 0, where a = -16, b = 64, and c = 80.
We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a).
Substituting the values for a, b, and c, we get:
t = (-64 ± √(64^2 - 4*(-16)*80)) / (2*(-16)).
Simplifying further:
t = (-64 ± √(4096 + 5120)) / (-32).
t = (-64 ± √9216) / (-32).
t = (-64 ± 96) / (-32).
We have two possible values for t:
t1 = (-64 + 96) / (-32) = 32 / (-32) = -1.
t2 = (-64 - 96) / (-32) = -160 / (-32) = 5.
Therefore, the object hits the ground after approximately 5 seconds.