the height, in feet, of a free-falling object t seconds after being dropped from an initial height s can be found using h=-16t^2+s. An object is dropped from a helicopter 784 feet above the ground. How long will it take the object to land on the ground?
how do you know where to put what??
a. 7s
b. 9s
c. 6s
d. 8.5s
h = Vo(t) + g(t^2) where h = the initial height, 784 feet, Vo = the initial velocity = 0, g = the acceleration due to gravity, 32 fps^2 and t = the time to fall from h = 784 to h = 0.
Therefore, 784 = (0)t + 32(t^2).
Solve for t.
ok I still do not understand this is what I got so far from the word problem above.
784=-16(0t^2)+s
784=-16+s
784+16=t^2+s
800
and the square root of 800 is 28
am I doing this right?
ok I still do not understand this is what I got so far from the word problem above.
784=-16(0t^2)+s
784=-16+s
784+16=t^2+s
800
and the square root of 800 is 28
am I doing this right?
Using your terminology:
s = the altitude of the helicopter
t = the time from the release of the object from the helicopter
h = the height of the object above the ground after t seconds
h = -16(t^2) + s where -16 is the coefficient of t^2
You have altered the real equation to read h = -16 + t^2 + s, and then ignored the term "s".
Remember that s = the altitude of the helicopter = 784 feet and h = the height of the object (the ground)after "t" seconds or h = "0".
Then 0 = -16(t^2) + 784 or
16(t^2) = 784 or t^2 = 49 making t = 7 seconds.
To find out how long it will take for the object to land on the ground, we need to determine when the height of the object (h) is equal to zero. In the given equation h = -16t^2 + s, h represents the height of the object at time t, and s represents the initial height, which in this case is 784 feet.
Since we want to know when the height is zero, we can set h = 0 and substitute the given values: 0 = -16t^2 + 784.
Now we have a quadratic equation in the form of at^2 + bt + c = 0, where a = -16, b = 0, and c = 784. To solve for t, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a).
Plugging in the values, we get t = (± √((-16)*(0)^2 - 4*(-16)*784)) / (2*(-16)).
Simplifying, we have t = (± √(0 + 50176)) / (-32).
Calculating further, t = (± √(50176)) / (-32).
Taking the square root of 50176, we find that √(50176) = ± 224.
Plugging in this value, we have t = (±224) / (-32).
Simplifying further, t = ±7.
Since time cannot be negative in this context, we can disregard the negative solution. Therefore, the answer is t = 7.
So, the object will take 7 seconds to land on the ground.
Therefore, the correct option is a. 7s.