In Mostar, Bosnia, the ultimate test of a young man's courage once was to jump off a 400 year old bridge into the River Neretva, 23 m below the bridge. (a) How long did the jump last? (b) How fast was the diver traveling upon impact with the river? (c)if the speed of sound in air is 340 m/s, how long after the diver took off did a spectator on the bridge hear the splash?
Work: I am not sure if I am doing this correctly.
(a) (v2)^2=(0)^2+2(9.8 m/s^2)(23m)
v2 = 21.2m/s
9.8m/s^2=(21.2m/s)/t
t=2.16s
(b) I don't know
(c) (340m/s)(2.16s)=157m
(b)-->(a)21.2m/s
(c)(23m)(s/340m)=0.0068s
plus 2.16 from (a)
To calculate the answer to part (b), we need to use the equation for motion with constant acceleration:
v = u + at
Where:
v is the final velocity
u is the initial velocity (which is 0 in this case)
a is the acceleration (which is due to gravity and is -9.8 m/s^2)
t is the time
We know that the final velocity (v) is what we need to find. The initial velocity (u) is 0 because the diver starts from rest. The acceleration (a) is -9.8 m/s^2 because it's directed towards the ground. And the time (t) is the time it takes for the diver to fall, which we calculated as 2.16 seconds in part (a).
So, using the equation, we have:
v = 0 + (-9.8 m/s^2) * 2.16 s
v = -21.168 m/s
The negative sign indicates that the velocity is directed downwards. Since velocity is a vector quantity, we only need to consider the magnitude of the velocity and ignore the direction.
Therefore, the diver was traveling at a speed of approximately 21.168 m/s upon impact with the river.
Now, let's move on to part (c). We are given the speed of sound in air, which is 340 m/s. We want to find how long after the diver took off did a spectator on the bridge hear the splash.
Since sound travels at a constant speed, we can use the equation:
distance = speed * time
In this case, the distance is given as 157 m and the speed is 340 m/s. We need to find the time (t).
Rearranging the equation, we have:
time = distance / speed
Plugging in the values, we get:
time = 157 m / 340 m/s
time ≈ 0.461 s
Therefore, a spectator on the bridge would hear the splash approximately 0.461 seconds after the diver took off.
Let's go through each part of the problem step-by-step.
(a) To find out how long the jump lasted, we can use the equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
In this case, the initial velocity (u) is 0 m/s (since the diver starts from rest), the acceleration (a) is 9.8 m/s^2 (acceleration due to gravity), and the distance (s) is 23 m.
Plugging in the values, we have:
v^2 = 0^2 + 2 * 9.8 * 23
v^2 = 0 + 450.8
v^2 = 450.8
To find v, we take the square root of both sides:
v = √450.8
v ≈ 21.24 m/s
So, the diver's final velocity upon impact with the river (or just before) is approximately 21.24 m/s.
(b) To find out how fast the diver was traveling upon impact with the river, we have already calculated the answer in part (a). The diver's speed upon impact is approximately 21.24 m/s.
(c) To calculate the time it takes for the sound of the splash to reach a spectator on the bridge, we can use the speed of sound in air, which is given as 340 m/s.
We know the time it took for the jump (2.16 s from part a), so we multiply this time by the speed of sound:
t = 2.16 s
d = v * t
d = 340 * 2.16
d ≈ 734.4 m
Therefore, the spectator on the bridge would hear the splash approximately 734.4 m away from the diver, which is about 0.7344 km.
I hope this helps! Let me know if you have any further questions.