An object is launched at 9.8 meters per second from a 73.5-meter tall platform. The object's height s (in meters) after t seconds is given by the equation s(t)= -4.9t(-4.9t) - 9.8t = 73.5. When does the object strike the ground?
I literally have no idea where to start
You first multiply -4.9t by -4.9t.
Then you subtract 73.5 from both sides and from there you use the quadratic formula.
thanks
To find when the object strikes the ground, we need to determine the value of t when the height (s) is equal to zero.
Let's start by substituting 0 for s in the equation and solve for t:
0 = -4.9t^2 - 9.8t + 73.5
This equation is in the form of a quadratic equation (ax^2 + bx + c = 0), where a = -4.9, b = -9.8, and c = 73.5.
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values from the equation, we have:
t = (-(-9.8) ± √((-9.8)^2 - 4(-4.9)(73.5))) / (2(-4.9))
Simplifying this equation, we get:
t = (9.8 ± √(9.8^2 + 4(4.9)(73.5))) / (-9.8)
Now, we can calculate the value of t using a calculator:
t ≈ (-9.8 ± √(96.04 + 1141.16)) / (-9.8)
t ≈ (-9.8 ± √1237.2) / (-9.8)
t ≈ (-9.8 ± 35.17) / (-9.8)
This gives us two possible values for t:
t ≈ (-9.8 + 35.17) / (-9.8) ≈ 2.002
t ≈ (-9.8 - 35.17) / (-9.8) ≈ 7.006
Since we are looking for the time when the object strikes the ground, we discard the negative value. Therefore, the object will strike the ground approximately 2.002 seconds after it is launched.