Answer:
2.8 m 7.4 m/s
Explanation:
write all the values then use the equations of motion to find the distance and speed. please see attached photo
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Answer:
We kindly invite you to read carefully the explanation and check the image attached below.
Explanation:
According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:
(1)
Where:
- Initial velocity, measured in meters per second.
- Final velocity, measured in meters per second.
- Acceleration, measured in meters per square second.
- Initial time, measured in seconds.
- Final time, measured in seconds.
Now we obtain the kinematic equations for thrust and free fall stages:
Thrust (, , , )
(2)
Free fall (, , , )
(3)
Now we created the graph speed-time, which can be seen below.
consider the motion in x-direction
= initial velocity in x-direction = ?
X = horizontal distance traveled = 100 m
= acceleration along x-direction = 0 m/s²
t = time of travel = 4.60 sec
Using the equation
X = t + (0.5) t²
100 = (4.60)
= 21.7 m/s
consider the motion along y-direction
= initial velocity in y-direction = ?
Y = vertical displacement = 0 m
= acceleration along x-direction = - 9.8 m/s²
t = time of travel = 4.60 sec
Using the equation
Y = t + (0.5) t²
0 = (4.60) + (0.5) (- 9.8) (4.60)²
= 22.54 m/s
initial velocity is given as
= sqrt(()² + ()²)
= sqrt((21.7)² + (22.54)²) = 31.3 m/s
direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg