Standing side by side, you and a friend step off a bridge at different times and fall for 1.41 s to the water below. Your friend goes first, and you follow after she has dropped a distance of 1.93 m. (a)How long does it take your friend to fall 1.93 m? (b)What is the separation between the two of you when your friend hits the water?
t = √(2s/a) = √(2*1.93/9.8)
separation is thus 1/2 a(1.41^2 - (1.41-t)^2)
To solve this problem, we can use kinematic equations for free-falling motion.
(a) The equation we'll use is:
d = 0.5 * g * t^2
where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
For your friend, we have:
1.93 m = 0.5 * 9.8 m/s^2 * t^2
Rearranging the equation, we can find the time:
t^2 = (2 * 1.93 m) / 9.8 m/s^2
t^2 = 0.3949
t ≈ 0.63 s
So it takes your friend approximately 0.63 seconds to fall 1.93 meters.
(b) To find the separation between the two of you when your friend hits the water, we need to determine how far you've fallen in the same 1.41 seconds.
Using the same formula, d = 0.5 * g * t^2, we have:
d = 0.5 * 9.8 m/s^2 * (1.41 s^2)
d ≈ 9.8 m
Therefore, the separation between you and your friend when she hits the water is approximately 9.8 meters.
To solve this problem, we can use the equations of motion for free fall. Let's start by calculating the time it takes for your friend to fall 1.93 m. We can use the equation:
h = (1/2) * g * t^2
where:
h is the distance fallen (1.93 m),
g is the acceleration due to gravity (9.8 m/s^2),
and t is the time.
Rearranging the equation, we have:
t^2 = (2 * h) / g
Substituting the given values, we can calculate t:
t^2 = (2 * 1.93) / 9.8
t^2 = 0.3939
t ≈ 0.6276 seconds (rounded to four decimal places)
Therefore, it takes your friend approximately 0.6276 seconds to fall 1.93 m.
Now, let's calculate the separation between the two of you when your friend hits the water. Since you begin falling after your friend has dropped a distance of 1.93 m, your initial separation is also 1.93 m. We can use the equation:
s = ut + (1/2) * g * t^2
where:
s is the separation,
u is the initial velocity (which is 0 m/s since you start from rest),
t is the time, and
g is the acceleration due to gravity (9.8 m/s^2).
Substituting the values:
s = (0 * 0.6276) + (1/2) * 9.8 * 0.6276^2
s = (0) + (1/2) * 9.8 * 0.3939
s ≈ 1.9282 meters (rounded to four decimal places)
Therefore, the separation between the two of you when your friend hits the water is approximately 1.9282 meters.