A rocket fires from rest with an upward acceleration of 35 m/s2 for 2 seconds. After this time, the engine shuts off and the rocket freely falls back to the earth. The maximum height is 314 and the height the engine shuts off is 70 m. What is the amount of time the rocket is in the air?
the g in this instance is 10 m/s ^2
first step: h = 1/2 at^2 = 1/2 * 35 * 2^2 = 70 m
at this point, the velocity v = 35*2 = 70 m/s
so now, after t more seconds,
v = 70 - 10t
since v=0 at maximum height, we have t=7 when h = 314
so now we are at t=9
from this point after t more seconds, h=314 -5t^2 and h=0 when t=7.92
so total flight time is about 2+7+8 = 17 s
If it accelerates from rest at 35 m/s^2 for 2 seconds it reaches
(1/2) a t^2 = (35/2)(4) = 70 meters
so agree but wonder why they said it.
the engine shuts off at 70 meters
The rocket then coasts up to 314meters
What is initial speed at 70 m?
Vi = a t = 35 * 2 = 70 m/s at 70 meters and 2 seconds
now it coasts up an additional 314-70 = 244meters
v = Vi - 10 t = 70 - 10 t
v = 0 when t = 10 so 12 seconds in the air so far
now it falls 314 meters
314 = (10/2) t^2
t^2 = 62.8
t = 7.92 seconds
so total time = 12 + 7.92 = about 20 seconds
whoops arithmetic !
v = Vi - 10 t = 70 - 10 t
v = 0 when t = 7 so 9 seconds in the air so far
now it falls 314 meters
314 = (10/2) t^2
t^2 = 62.8
t = 7.92 seconds
so total time = 9 + 7.92 = about 17 seconds
To find the amount of time the rocket is in the air, we need to consider two parts: the time it takes for the rocket to reach its maximum height during the upward acceleration, and the time it takes for the rocket to fall back to the ground.
Let's break down the problem step by step:
Step 1: Calculate the time it takes for the rocket to reach its maximum height during the upward acceleration.
We know that the rocket's upward acceleration is 35 m/s^2, and it fires for 2 seconds. Using the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity (which is 0 m/s since the rocket starts from rest)
a = acceleration
t = time
Plugging in the values:
v = 0 + (35 m/s^2) * (2 s)
v = 70 m/s
Step 2: Determine the time it takes for the rocket to fall back to the ground.
The rocket will freely fall back to the Earth, which means it will experience the acceleration due to gravity, which is approximately 9.8 m/s^2. The height at which the engine shuts off is 70 m, and we need to find the time it takes for the rocket to reach the ground from this height.
Using the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s since the rocket hits the ground)
u = initial velocity
a = acceleration due to gravity (-9.8 m/s^2, considering downward direction)
s = distance (which is 70 m since the rocket starts from this height)
Plugging in the values:
0^2 = u^2 + 2 * (-9.8 m/s^2) * (70 m)
0 = u^2 - 1372 m^2/s^2
Solving this quadratic equation, we find:
u^2 = 1372 m^2/s^2
u ≈ 37.06 m/s
Since we are interested in the time taken, we will consider the positive value:
u = 37.06 m/s
Using the equation of motion again:
v = u + at
0 = 37.06 m/s + (-9.8 m/s^2) * t
Solving for t, we find:
-37.06 m/s = -9.8 m/s^2 * t
t ≈ 3.78 s
Step 3: Find the total time of flight.
To find the total time of flight, we need to add the time taken during upward acceleration to the time taken during free fall.
Total time = time taken during upward acceleration + time taken during free fall
Total time = 2 s + 3.78 s
Total time ≈ 5.78 s
So, the amount of time the rocket is in the air is approximately 5.78 seconds.