A rocket is launched straight up with a constant acceleration. Three seconds after lift off, a small part of the rocket falls off the side. The small part hits the ground 7 seconds later. Ignoring air resistance, what was the acceleration of the rocket?
Initial velocity of fragment, at release
Vo = 3 a
Height where fragment is released:
H = (a/2)t^2 = (9/2)a
Since Time to reach ground = 7s:
y(t=7) = H +Vo*7 -(g/2)*49 = 0
(9/2)a + 21a -24.5g = 0
a = (24.5/25.5) g = 9.4 m/s^2
To find the acceleration of the rocket, we can use the kinematic equation for an object in vertical motion, which is given by:
s = ut + 1/2at^2
where:
s = vertical distance traveled (in this case, the distance the small part of the rocket falls)
u = initial velocity in the vertical direction (0, as the small part fell off the side of the rocket)
t = time taken (7 seconds, as given in the question)
a = acceleration (we need to solve for this)
In this case, we need to consider the time taken by the small part of the rocket, starting from the moment it fell off the side until it hit the ground. If the rocket was launched straight up and the small part fell off the side, it means that the vertical motion of the small part is independent of the rocket's motion. Therefore, we can treat the motion of the small part as a separate object with its own acceleration.
Let's denote the distance the small part falls off the side as d (which is the same as the distance it travels vertically), the time it takes to fall off the side and hit the ground as t_s, and the acceleration of the small part as a_s.
From the given information, we can deduce the following:
- The small part takes t_s = 7 seconds to hit the ground.
- The small part fell off the side 3 seconds after lift off, so it took a total of (3 + t_s) seconds to fall from the rocket.
Using the kinematic equation, we can determine the value of acceleration for the small part as follows:
d = 0 + 1/2 a_s (3 + t_s)^2 (since the initial velocity is 0)
d = 1/2 a_s (3 + t_s)^2
Plugging in the values of d and t_s:
d = 1/2 a_s (3 + 7)^2
d = 1/2 a_s (10)^2
d = 50 a_s
Now, we have an expression for the distance traveled by the small part in terms of its acceleration.
Next, we'll consider the rocket. We know that the rocket was launched straight up with a constant acceleration, denoted as a. Since the small part fell off the side, its acceleration is different from the rocket's acceleration.
As mentioned earlier, the vertical motion of the small part is independent of the rocket's motion. Therefore, the distance traveled by the small part is equal to the distance the rocket has traveled at the time the small part falls off the side.
So, we have:
50 a_s = ut + 1/2a(3 + t_s)^2
Since the initial velocity of the rocket in the vertical direction is 0, this simplifies to:
50 a_s = 1/2a(3 + t_s)^2
Now, we substitute the values we have:
50 a_s = 1/2a(3 + 7)^2
50 a_s = 1/2a(10)^2
50 a_s = 50 a
We can cancel out the factor of 50 on both sides:
a_s = a
Therefore, the acceleration of the small part is equal to the acceleration of the rocket. Hence, the acceleration of the rocket is the same as the acceleration of the small part, which we can calculate using the given data.