To find out when the ball hits the ground, we need to find the time at which the height h is equal to 0.
So we set h = 0:
0 = 151 - 15t - 16t^2
Rearranging the equation:
16t^2 + 15t - 151 = 0
Now we can solve this quadratic equation using the quadratic formula:
t = (-15 ± √(15^2 - 4(16)(-151))) / (2(16))
t = (-15 ± √(225 + 9672)) / 32
t = (-15 ± √9897) / 32
Since we're looking for the time after which the ball hits the ground, we only consider the positive value of t:
t ≈ 3.5 seconds
Therefore, the ball hits the ground approximately 3.5 seconds after it is thrown.
A ball is thrown from a height of 151 feet with an initial downward velocity of 15 ft/s. The ball's height h (in feet) after t seconds is given by the following
h = 151-15t-16t^2
How long after the ball is thrown does it hit the ground?
1 answer