A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc
1 answer:
Answer:
-1.43 m/s relative to the shore
Explanation:
Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:
where are the mass of the swimmer and raft, respectively.
are the velocities of the swimmer and the raft after the run, respectively. We can solve for
So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore
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Answer:
a) The takeoff speed is 10 m/s.
b) The maximum height above the ground is 1.2 m.
Explanation:
The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v =(v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t".
x0 = initial horizontal position.
v0 = initial velocity.
α = jumping angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
v = velocity vector at time "t"
a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:
r final = (8.7 m, 0 m)
Then at final time:
8.7 m = x0 + v0 · t · cos α
0 m = y0 + v0 · t · sin α + 1/2 · g · t²
(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:
8.7 m = v0 · t · cos α
Solving for "v0":
8.7 m /(t · cos α) = v0
Replacing v0 in the equation of the y-component, we can obtain the final time:
0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)
- 8.7 m · tan 29° / -4.9 m/s² = t²
t = 0.99 s
Now, we can calculate the initial speed:
8.7 m /t · cos α = v0
v0 = 8.7 m / (0.99 s · cos 29°)
<u>v0 = 10 m/s</u>
The takeoff speed is 10 m/s
b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:
0 = v0 · sin α + g · t
Solving for "t":
-v0 · sin α / g = t
t = - 10 m/s · sin 29° / 9.8 m/s²
t = 0.49 s
Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.
Now, let´s find the height of the kangaroo at that time:
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²
<u>y = 1.2 m</u>
The maximum height above the ground is 1.2 m.
