Madison is standing on the roof of a building that is 350 feet above the floor. She throws a ball straight up with a velocity of 20 feet per second. How long will the ball be in the air before it hits the ground?
first we observe that the motion of the ball is uniformly accelerated motion (because of acceleration due to gravity while it's in the air), thus we use this formula to find time:
h = ho + vo*t - (1/2)*g*t^2
where
ho = initial height
vo = initial velocity
t = time
g = acceleration due to gravity (9.8 m/s^2 = 32 ft/s^2)
making the ground as the reference or the origin (thus at this point, the height = h = 0), ho is thus equal to 350
substituting,
0 = 350 + 20*t - 1/2*32*t^2
0 = 350 + 20t - 16t^2
since it's a quadratic equation, we either factor it or use quadratic formula,
t = [-b +- sqrt(b^2 - 4ac)]/(2a)
subsituting,
t = [-20 +- sqrt(20^2 - 4(-16)(350))]/(2(-16))
t = [-20 +- 150.997]/(-32)
t = -4.09
t = 5.34
since time is always positive, we get the positive root:
t = 5.34 s
hope this helps~ :)
It depends on many factors not said in the problem. How high does the ball go before it starts coming down, for example.
The problem is not solvable.
Ok. Someone else was smarter than I. I should have waited, I guess.
thanks
To find out how long the ball will be in the air before hitting the ground, we can use the kinematic equation for vertical displacement:
h = vi * t - (1/2) * g * t^2
Where:
h = vertical displacement (change in height)
vi = initial velocity (20 feet per second)
t = time in seconds
g = acceleration due to gravity (32 feet per second^2)
Given that Madison is standing 350 feet above the ground, the initial height is the vertical displacement, which is -350 feet (negative because the ball is moving downward).
Therefore, we can rewrite the equation as:
-350 = 20 * t - (1/2) * 32 * t^2
Simplifying the equation:
-350 = 20t - 16t^2
Rearranging the terms and setting the equation equal to zero:
16t^2 - 20t - 350 = 0
Now, we can solve the quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -20, and c = -350. Plugging in the values:
t = (20 ± √((-20)^2 - 4 * 16 * -350)) / (2 * 16)
Simplifying further:
t = (20 ± √(400 + 22400)) / 32
t = (20 ± √22800) / 32
Now, we can calculate the two possible values of t by substituting in the ± sign:
t = (20 + √22800) / 32
t = (20 - √22800) / 32
This gives us two values of t. However, one value will be negative and does not make sense in this context since time cannot be negative. So we will take the positive value:
t ≈ (20 + √22800) / 32
Calculating this value using a calculator, we get:
t ≈ 4.53 seconds
Therefore, the ball will be in the air for approximately 4.53 seconds before hitting the ground.