a model rocket rises with constant accerleration to a height of 3.2 m, at which point its speed is 26.0 m/s. (a) how much time does it take for the rocket to reach this height? (b) what was the magnitude of the rocket's acceration? (c) find the height and speed of the rocket 0.10 s after launch.
Let's solve the problem step by step:
(a) To find the time it takes for the rocket to reach a height of 3.2 m, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity = 26.0 m/s
u = initial velocity = 0 m/s (since the rocket starts from rest)
a = acceleration (constant)
s = displacement = 3.2 m
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the given values, we get:
a = (26.0^2 - 0^2) / (2 * 3.2)
a = (676 - 0) / 6.4
a = 676 / 6.4
a ≈ 105.625 m/s^2
(b) The magnitude of the rocket's acceleration is approximately 105.625 m/s^2.
(c) To find the height and speed of the rocket 0.10 s after launch, we can use the following equations:
s = ut + (1/2)at^2
v = u + at
where:
t = time = 0.10 s
u = initial velocity = 0 m/s (since the rocket starts from rest)
a = acceleration = 105.625 m/s^2
Let's calculate the height first:
s = ut + (1/2)at^2
s = 0 + (1/2)(105.625)(0.10)^2
s = 0.5 * 105.625 * 0.01
s ≈ 0.53125 m
The height of the rocket 0.10 s after launch is approximately 0.53125 m.
Now, let's calculate the speed:
v = u + at
v = 0 + (105.625)(0.10)
v ≈ 10.5625 m/s
The speed of the rocket 0.10 s after launch is approximately 10.5625 m/s.
To solve this problem, we can use the equations of motion for an object moving with constant acceleration.
(a) To find the time it takes for the rocket to reach a height of 3.2 m, we can use the equation:
s = ut + (1/2)at^2
Where:
s = displacement or height (3.2 m)
u = initial velocity (0 m/s since the rocket starts from rest)
a = constant acceleration
t = time
Rearranging the equation, we get:
t^2 = (2s)/a
Substituting the given values, we have:
t^2 = (2 * 3.2) / a
To solve for t, we need to know the value of acceleration (a).
(b) To find the magnitude of the rocket's acceleration, we can use the equation:
v = u + at
Where:
v = final velocity (26.0 m/s)
u = initial velocity (0 m/s)
a = constant acceleration
Rearranging the equation, we get:
a = (v - u) / t
Substituting the given values:
a = (26.0 - 0) / t
Now, we have two equations with two unknowns (t and a). We can solve them simultaneously to find the values.
(c) To find the height and speed of the rocket 0.10 s after launch, we can use the equations of motion again. Since the rocket's acceleration is constant, we can use the equations:
s = ut + (1/2)at^2
v = u + at
Given that t = 0.10 s, we can substitute this value into the equations to find the height (s) and speed (v) of the rocket.
Now, let's solve the equations to find the answers!