A toy rocket is launched upwards from rest. The rocket engine will cause the rocket to accelerate at 25 m/s^2 for 3.0 seconds.

a) How fast is the rocket moving aft the end of the 3 seconds?

b) What is the maximum height reached by the rocket?

c) What is the total time the rocket is in the air?

User Icon for Explain Bot Explain Bot answered
1 year ago

To find the answers to these questions, we can use equations of motion.

a) How fast is the rocket moving after the end of the 3 seconds?

To determine the velocity of the rocket after 3 seconds, we can use the equation:

v = u + at

where:
v = final velocity
u = initial velocity (which is 0 in this case)
a = acceleration (given as 25 m/s^2)
t = time (3.0 seconds)

Substituting the known values into the equation:

v = 0 + (25 m/s^2)(3.0 s)

v = 75 m/s

Therefore, the rocket is moving at a velocity of 75 m/s after 3 seconds.

b) What is the maximum height reached by the rocket?

To find the maximum height, we need to determine the time it takes for the rocket to reach its highest point. The time taken to reach the maximum height is half of the total time in the air, as the ascent and descent times are equal.

Given that the rocket is launched from rest and reaches its maximum height, we can use the equation:

v² = u² + 2as

where:
v = final velocity (which is 0 at the highest point)
u = initial velocity (which is 0 in this case)
a = acceleration (given as -9.8 m/s^2, as gravity acts in the opposite direction)
s = displacement

Rearranging the equation, we get:

s = (v² - u²) / (2a)

Substituting the known values:

s = (0 - 0) / (2 * -9.8 m/s^2)

s = 0 m

Therefore, the maximum height reached by the rocket is 0 meters. This implies that the rocket did not achieve any vertical displacement.

c) What is the total time the rocket is in the air?

To find the total time the rocket is in the air, we consider the ascent and descent phases separately.

During the ascent phase, the rocket experiences an upward acceleration of 25 m/s². Using the following equation:

v = u + at

where:
v = final velocity (which is 0 at the topmost point)
u = initial velocity (which is 0 in this case)
a = acceleration (given as 25 m/s²)
t = time taken during ascent

Rearranging the equation, we get:

t = (v - u) / a

Substituting the known values:

t = (0 - 0) / 25 m/s²

t = 0 s

During the descent phase, the rocket experiences a downward acceleration due to gravity of -9.8 m/s². Similarly, we can use the equation to calculate the time taken:

t = (v - u) / a

Using the same values as before:

t = (0 - 0) / (-9.8 m/s²)

t = 0 s

Since both the ascent and descent times are 0 seconds, the total time the rocket is in the air is also 0 seconds.