A certain freely falling object, released from rest, requires 1.40 s to travel the last 20.5 m before it hits the ground.
(a) Find the velocity of the object when it is 20.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.)
m/s
(b) Find the total distance the object travels during the fall.
m
Oh, falling objects can be quite hilarious! Let's solve this problem with a touch of humor.
(a) To find the velocity of the object when it is 20.5 m above the ground, we can use the kinematic equation: vf = vi + at.
First, let's assume the positive direction is upward. Since the object is falling downward, the acceleration due to gravity should be negative. So, a = -9.8 m/s² (because gravity has a way of bringing us down, literally!).
We know that the object is released from rest, so the initial velocity (vi) is 0 m/s. The time it takes to travel the last 20.5 m is given as 1.40 s.
Using the equation vf = vi + at, we can plug in the values:
vf = 0 + (-9.8)(1.40)
vf = -13.72 m/s
So, the velocity of the object when it is 20.5 m above the ground is -13.72 m/s. And by the way, the negative sign just means that gravity is pulling the object down, as it often does in our lives when we have high expectations!
(b) To find the total distance the object travels during the fall, we can use the equation s = vit + 1/2at².
Since the object is falling, we can use the equation s = 1/2gt², where g is the acceleration due to gravity. Plugging in the values:
s = 1/2(-9.8)(1.40)²
s = 10.92 m
So, the total distance the object travels during the fall is 10.92 m. It seems like the object took a little detour on its way down, but hey, who doesn't like a scenic route?
To solve this problem, we can use the equations of motion for a freely falling object under the influence of gravity. The two equations we need are:
1. vf = vi + gt, where vf is the final velocity, vi is the initial velocity (which in this case is 0 m/s since the object is released from rest), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
2. d = vi * t + (1/2) * g * t^2, where d is the distance traveled during the fall.
Let's start with part (a) and find the velocity of the object when it is 20.5 m above the ground.
Using equation 1, we have:
vf = 0 + (9.8 m/s^2) * 1.40 s
vf = 13.72 m/s
Since the object is moving downward, we assign a negative sign to the velocity. Therefore, the velocity of the object when it is 20.5 m above the ground is -13.72 m/s.
For part (b), we need to find the total distance the object travels during the fall.
To find the total distance, we first find the distance traveled during the last 20.5 m using equation 2.
20.5 m = 0 * t + (1/2) * (9.8 m/s^2) * (1.40 s)^2
20.5 m = 0 + 12.92 m/s^2 * (1.96 s^2)
20.5 m = 25.35 m
Now, we need to find the distance traveled from the release point to the 20.5 m mark. Since the object is in free fall, the time it takes to reach the 20.5 m mark is the same as the total time it takes to hit the ground. Therefore, we just need to double the time given.
t_total = 1.40 s + 1.40 s
t_total = 2.80 s
Now we can find the total distance traveled using equation 2:
d_total = 0 * (2.80 s) + (1/2) * (9.8 m/s^2) * (2.80 s)^2
d_total = 0 + 12.92 m/s^2 * (7.84 s^2)
d_total = 101.248 m
Therefore, the total distance traveled by the object during the fall is 101.248 m.
acceleration during last 20.5 m ... 1.40 s * 9.81 m/s^2 = 13.7 m/s
average velocity during last 20.5 m ... 20.5 m / 1.40 s = 14.6 m/s
(velocity at 20.5 m) + (impact velocity) = 2 * average = 29.2 m/s
impact velocity = velocity at 20.5 m + 13.7 m/s
2 (velocity at 20.5 m) + 13.7 m/s = 29.2 m/s
... the velocity is negative (downward)
impact velocity / g = flight time
1/2 * g * (flight time)^2 = total fall
a. V = Vo + g*t = 0 + 9.8*1.4 = 13.72 m/s.
b. h = ho - 0.5g*t^2 = 20.5.
ho - 4.9*1.4^2 = 20.5,
ho = 20.5 + 4.9*1.4^2 = 30.1 m. = Initial ht. above gnd. = Total distance.
V^2 = Vo^2 + 2g*d = 0 + 19.6(30.1-20.5) = 188.16,
V = 13.72