A rock is dropped from the top of a 400-foot-tall building. Its distance s (in feet) from the top of the building after t seconds is s(t) = 16t^2. How many seconds will it take the rock to hit the ground?
hit the ground == drop 400 ft. So,
16t^2 = 400
Now just solve for t.
retga
To find the amount of time it takes for the rock to hit the ground, we need to determine when its distance from the top of the building, s(t), becomes zero.
The given equation for the distance of the rock from the top of the building is s(t) = 16t^2. For the rock to hit the ground, its distance, s(t), should be zero.
Setting s(t) to zero, we have:
0 = 16t^2
To solve this quadratic equation, we can factor out a common factor of 16:
0 = 16(t^2)
Now, we have a product of two factors equating to zero. According to the zero product property, if the product of two factors is zero, then at least one of the factors must be zero.
So, we set each factor equal to zero and solve for t:
t^2 = 0
Since the square of a number is only equal to zero when the number itself is zero, we have:
t = 0
Therefore, the rock will hit the ground at t = 0 seconds.
However, it's important to note that the initial conditions of the problem are not mentioned. Usually, an object is dropped from rest, which means its initial velocity is zero. In that case, we can assume the initial conditions are consistent with this standard scenario.