During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.50 m/s^2. When it is 255 m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored).
What total distance did the canister travel between its release and its crash onto the launch pad?
IT has an initial velocity, and height.
so find the initial velocity...
vf^2=2ad
Vf= sqrt(2*.350*255)
so how much higher does it go?
1/2 m v^2=mgh
solve for h additional, and then it falls h + 255m to the ground.
d=2*habove+255
To find the total distance traveled by the canister between its release and its crash onto the launch pad, we need to calculate two distances: the distance traveled while accelerating and the distance traveled while falling back down to the launch pad.
First, let's calculate the distance traveled while accelerating. We can use the kinematic equation:
s = ut + (1/2)at^2
where,
s = distance traveled while accelerating,
u = initial velocity (which is 0 in this case, as the canister starts from rest),
a = acceleration (3.50 m/s^2 in this case),
t = time.
We need to find the time it takes for the canister to reach a height above the launch pad of 255 m. We can use another kinematic equation:
v^2 = u^2 + 2as
where,
v = final velocity (which is 0 when the canister reaches its highest point),
u = initial velocity (0 m/s),
a = acceleration (3.50 m/s^2),
s = displacement (255 m).
Rearranging the equation, we get:
0 = 0^2 + 2(3.50)(-255)
0 = 0 + (-1785)
1785 = 2(3.50)s
s = 1785 / (2 * 3.50)
s = 255 m
Now that we know the canister reaches a height of 255 m, we can calculate the time it takes to reach that height using the first kinematic equation:
s = ut + (1/2)at^2
255 = 0t + (1/2)(3.50)t^2
255 = (1.75)t^2
t^2 = 255 / 1.75
t^2 = 145.71
t ≈ √145.71
t ≈ 12.07 s
Therefore, the time taken to reach a height of 255 m is approximately 12.07 seconds.
Now, let's calculate the distance traveled while falling back down to the launch pad. When the canister is in free fall, the only force acting on it is gravity, resulting in a constant acceleration of 9.8 m/s^2. The distance traveled while falling can be calculated using the kinematic equation:
s = ut + (1/2)at^2
where,
s = distance traveled while falling,
u = initial velocity (which is 0 when the canister is released),
a = acceleration (9.8 m/s^2),
t = time.
Since the canister is in free fall, we can use the same time it took for the canister to reach a height of 255 m (approximately 12.07 seconds).
Plugging in the values, we get:
s = 0(12.07) + (1/2)(9.8)(12.07)^2
Simplifying,
s = (1/2)(9.8)(12.07)^2
s ≈ 712.05 m
Therefore, the distance traveled while falling back down to the launch pad is approximately 712.05 meters.
To find the total distance traveled, we simply add the distance traveled while accelerating (255 m) and the distance traveled while falling (712.05 m):
Total distance = 255 m + 712.05 m
Total distance ≈ 967.05 m
Hence, the total distance traveled by the canister between its release and its crash onto the launch pad is approximately 967.05 meters.