Question

A ball is thrown upward with an initial velocity of 48 ft/sec from a height 640ft. It's height h, in feet, after t seconds is given by h(t)= -16t^2 +48t+640. After how long will the ball reach the ground?

The ball will reach the ground in ___seconds

Answer 1

To find when the ball will reach the ground, we need to find the time when the height is equal to zero. By setting the equation for height equal to zero, using the quadratic formula, and taking the positive solution, we find that the ball will reach the ground in approximately 3.79 seconds.

Step-by-step explanation:

To find out when the ball will reach the ground, we need to find the time when the height is equal to zero. We can do this by setting the equation for height, h(t), equal to zero and solving for t. The equation is: h(t) = -16t^2 + 48t + 640. So, we have: -16t^2 + 48t + 640 = 0. From here, we can use the quadratic formula to solve for t. The positive solution will give us when the ball reaches the ground.

Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), where a = -16, b = 48, and c = 640, we can plug in these values to find the value of t. Evaluating the equation using these values gives us t = 3.79 seconds and t = 0.54 seconds. Since the ball is thrown upward and then comes back down, the longer solution, t = 3.79 seconds, represents the time it takes for the ball to reach the ground.

Answer 2

We equate h(t) to 0 to find the time at which the ball reaches the ground.
16t² - 48t - 640 = 0
t² - 3t - 40 = 0
t = 8, t = -5
Time cannot be negative so second value neglected.
The ball will reach the ground in 8 seconds.