A model rocket blasts off and moves upward with an acceleration of 12 m/s^2 until it reaches a height of 28 m , at which point its engine shuts off and it continues its flight in free fall.
What is the speed of the rocket just before it hits the ground?
So there are two steps since two accelerations are involved.
Step 1: acceleration = +12 m/s/s
initial velocity, u=0
final velocity = v
distance travelled, S=28 m
using
V²-u²=2aS
we find
v²
=2aS+u²
=2*12*28+0
=4√42 m/s (upwards)
=25.92 m/s (upwards)
Step 2: free fall, a=-9.81 m/s/s
initial velocity, u = 4√42 m/s
distance travelled, S = -28 m
final velocity = v
Again, we use the formula
V²-u²=2aS
from which
v²
=2aS+u²
=2*(-9.81)*(-28)+(4√42)²
=1221.36
v=√1221.36
=34.95 m/s (downwards)
To find the speed of the rocket just before it hits the ground, we can use the equations of motion.
First, let's break the problem down into two parts:
1. The acceleration phase, when the rocket blasts off and moves upward with an acceleration of 12 m/s^2 until it reaches a height of 28 m.
2. The free fall phase, when the rocket continues its flight without any acceleration.
For the first part, we can use the following equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the rocket starts from rest (u = 0) and accelerates upward with an acceleration of 12 m/s^2, we can substitute the values into the equation:
v^2 = 0^2 + 2 * 12 * 28
v^2 = 0 + 672
v^2 = 672
Taking the square root of both sides to solve for v, we have:
v = √672
v ≈ 25.98 m/s
Therefore, the speed of the rocket just before it reaches a height of 28 m is approximately 25.98 m/s.
Now, for the second part, when the rocket is in free fall, it will continue to fall with a constant acceleration due to gravity (approximately 9.8 m/s^2) until it hits the ground. This means its speed will keep increasing at a rate of 9.8 m/s every second.
However, since we already know the final speed just before reaching the ground, we don't need to calculate it again. So, the speed of the rocket just before it hits the ground is approximately 25.98 m/s.
To determine the speed of the rocket just before it hits the ground, we can break down the problem into two parts: the upward motion of the rocket and its subsequent descent during free fall.
First, let's calculate the time it takes for the rocket to reach its peak height during the upward part of its motion. We can use the equation:
s = ut + 0.5at^2
where s is the displacement, u is the initial velocity (which is 0 m/s in this case), a is the acceleration, and t is the time.
Given:
s = 28 m
u = 0 m/s
a = 12 m/s^2
Rearranging the equation, we have:
t^2 = (2s) / a
t^2 = (2 * 28) / 12
t^2 = 56 / 12
t^2 = 4.67
Taking the square root of both sides, we get:
t = sqrt(4.67)
t ≈ 2.16 s
So, the time taken by the rocket to reach its peak height is approximately 2.16 seconds.
Next, let's determine the time it takes for the rocket to fall back down to the ground from its peak height. During free fall, the only force acting on the rocket is gravity, which causes it to accelerate downwards at 9.8 m/s^2 (approximately).
We can use the equation:
s = ut + 0.5at^2
But in this case, the initial velocity (u) is not zero, as the rocket starts falling from its peak height. We need to find the value of u.
We know:
s = 28 m (height reached by the rocket)
a = 9.8 m/s^2 (acceleration due to gravity)
t = 2.16 s (time from peak height to the ground)
Using the equation and substituting the known values, we have:
28 = u * 2.16 - (0.5 * 9.8 * 2.16^2)
Now, we solve this equation to find the initial velocity (u) of the rocket as it starts to fall. Once we have u, we can calculate the speed just before it hits the ground.
Let's solve the equation to find u.