a ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity 80 m/ sec .the distance s of the ball from the ground after t seconds is s= 96+80t-4.9 t^2.after how many seconds does the ball strike the ground
To find the time at which the ball strikes the ground, we need to determine when the distance from the ground is zero.
The given equation for the distance of the ball from the ground is s = 96 + 80t - 4.9t^2. We can set this equation equal to zero to represent the ball being on the ground:
0 = 96 + 80t - 4.9t^2
Now, we can solve this quadratic equation for t.
First, let's rearrange the equation:
4.9t^2 - 80t - 96 = 0
This quadratic equation can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 4.9, b = -80, and c = -96.
Plugging these values into the quadratic formula, we have:
t = (-(-80) ± √((-80)^2 - 4(4.9)(-96))) / (2(4.9))
Simplifying further:
t = (80 ± √(6400 + 1881.6)) / 9.8
Now, we can calculate the values in the square root:
t = (80 ± √(8281.6)) / 9.8
t = (80 ± 90.91) / 9.8
Now, we'll find the two possible values for t:
t1 = (80 + 90.91) / 9.8 ≈ 17.1 seconds
t2 = (80 - 90.91) / 9.8 ≈ -1.1 seconds
Since time cannot be negative in this context, we discard t2 = -1.1 seconds.
Therefore, the ball strikes the ground after approximately 17.1 seconds.
4r3435
you want height to be zero, so
0 = 96+80t -4.9t^2
using the formula ...
t = (-80 ± √8281.6)/-9.8
= appr 17.45 seconds or a negative