Question

A rocket accelerates straight up from the ground at 12.6 m/s^2 for 11.0 s. Then the engine cuts off and the rocket enters free fall. (a) Find its velocity at the end of its upward acceleration. (b) What maximum height does it reach? (c) With what velocity does it crash to Earth? (d) What's the total time from launch to crash?

Answer 1

The rocket's velocity at the end of its upward acceleration is 138.6 m/s. The maximum height it reaches is 1353.2 m. The rocket crashes to Earth with a velocity of -1.0 m/s.

Step-by-step explanation:

(a) To find the velocity at the end of the rocket's upward acceleration, we can use the formula:

v = u + at

Where:
u = initial velocity = 0 m/s (since the rocket starts from rest)
a = acceleration = 12.6 m/s2
t = time = 11.0 s

Substituting the values, we get:

v = 0 + 12.6 * 11.0 = 138.6 m/s

Therefore, the velocity at the end of its upward acceleration is 138.6 m/s.

(b) To find the maximum height the rocket reaches, we can use the second equation of motion:

s = ut + 0.5at2

Where:
s = distance
u = initial velocity
a = acceleration
t = time

In this case, we need to consider the time taken for both the upward acceleration and free fall.

For the upward acceleration:

u = 0 m/s (since the rocket starts from rest)
a = 12.6 m/s2
t = 11.0 s

Substituting the values, we get:

s1 = 0 * 11.0 + 0.5 * 12.6 * (11.0)2 = 388.5 m

For the free fall:

u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = ?

To find the time for free fall, we can use the equation:

u = at

Substituting the values, we get:

138.6 = -9.8t

Solving for t, we get:

t = -14.1 s

However, time cannot be negative in this case. So, we take the absolute value of t:

t = 14.1 s

Substituting the values in the equation for free fall distance, we get:

s2 = 138.6 * 14.1 + 0.5 * (-9.8) * (14.1)2 = 964.7 m

The maximum height reached by the rocket is s1 + s2 = 388.5 m + 964.7 m = 1353.2 m.

(c) To find the velocity at which the rocket crashes to Earth, we again consider the free fall phase. Using the equation:

v = u + at

Where:
u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = 14.1 s

Substituting the values, we get:

v = 138.6 - 9.8 * 14.1 = -1.0 m/s

The velocity at which the rocket crashes to Earth is -1.0 m/s. The negative sign indicates that the velocity is directed downward.

(d) The total time from launch to crash is the sum of the time for upward acceleration (11.0 s) and the absolute value of the time for free fall (14.1 s). Therefore, the total time is 11.0 s + 14.1 s = 25.1 s.

Answer 2

a) 138.6 m/s

b) 762.3 m

c) 122.3 m/s

d) 24.47

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

`v=u+at\\\Rightarrow v=0+12.6* 11\\\Rightarrow v=138.6 \ m/s`

Velocity at the end of its upward acceleration is 138.6 m/s

`s=ut+(1)/(2)at^2\\\Rightarrow s=0* t+(1)/(2)* 12.6* 11^2\\\Rightarrow s=762.3\ m`

Maximum height the rocket reaches is 762.3 m

`v^2-u^2=2as\\\Rightarrow v=√(2as-u^2)\\\Rightarrow v=√(2* 9.81* 762.3-0^2)\\\Rightarrow v=122.3\ m/s`

The velocity with which the rocket crashes to the Earth is 122.3 m/s

`s=ut+(1)/(2)at^2\\\Rightarrow 762.3=0* t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(762.3* 2)/(9.81)}\\\Rightarrow t=12.47\ s`

Total time from launch to crash is 12.47+11 = 24.47 seconds