Suppose an object is dropped from a height h0 above the ground. Then its height after t seconds is given by h=−16t^2+h0, where h is measured in feet. If a ball is dropped from 48 feet above the ground, how long does it take to reach ground level?
just plug in 48 for h0, and solve for t when h(t) = 0. Just the usual quadratic stuff.
To find out how long it takes for the ball to reach ground level, we need to set the height h to zero and solve for t in the equation h = -16t^2 + h₀.
Given that the ball is dropped from a height h₀ of 48 feet, the equation becomes:
0 = -16t^2 + 48
To solve this equation, we can rearrange it to isolate the variable t:
16t^2 = 48
Dividing both sides of the equation by 16:
t^2 = 3
Taking the square root of both sides:
t = ±√3
Since time cannot be negative, we only consider the positive square root:
t ≈ √3
Therefore, it takes approximately √3 seconds for the ball to reach ground level.
To find how long it takes for the ball to reach ground level, we need to find the value of t when the height (h) is equal to zero.
Given that the height of the ball after t seconds is given by the equation h = -16t^2 + h0, we can substitute h = 0 and h0 = 48 into the equation:
0 = -16t^2 + 48
Now, we can solve for t by rearranging the equation:
16t^2 = 48
Divide both sides of the equation by 16:
t^2 = 3
To solve for t, we take the square root of both sides:
t = √3
Therefore, it takes approximately √3 seconds for the ball to reach ground level.