An object that is projected straight downward
with initial velocity v feet per second travels a
distance s vt 16t
2
=+ , where t =time in seconds.
If Ramón is standing on a balcony 84 feet above
the ground and throws a penny straight down
with an initial velocity of 10 feet per second, in
how many seconds will it reach the ground?
8.4 seconds
To find out in how many seconds the penny will reach the ground, we need to set up an equation based on the given information.
The equation for the distance traveled by the object is: s(t) = -16t^2 + vt, where s(t) is the distance in feet and t is the time in seconds.
Given that the initial velocity, v, is 10 feet per second and the object starts 84 feet above the ground, we can substitute these values into the equation: s(t) = -16t^2 + 10t + 84.
Since we want to know when the penny reaches the ground, we can set s(t) equal to zero and solve for t: -16t^2 + 10t + 84 = 0.
First, let's rearrange the equation: -16t^2 + 10t + 84 = 0.
To solve the quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring may not be straightforward, so we will use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a.
For our equation, a = -16, b = 10, and c = 84. Substituting these values into the formula, we get: t = (-10 ± √(10^2 - 4(-16)(84))) / 2(-16).
Simplifying further: t = (-10 ± √(100 + 5376)) / -32.
Continuing to simplify: t = (-10 ± √(5476)) / -32.
Taking the square root: t ≈ (-10 ± 74) / -32.
Now, we have two solutions: t ≈ (-10 + 74) / -32 = 64 / -32 = -2, and t ≈ (-10 - 74) / -32 = -84 / -32 = 2.625.
Since time cannot be negative, the only valid solution is t ≈ 2.625 seconds.
Therefore, it will take approximately 2.625 seconds for the penny to reach the ground.