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Akimi4 [234]
3 years ago
13

As a science fair project, you want to launch an 800 g model rocket st raight up and hit a horizontally moving target as it pass

es 30 m above the launch point. The rocket engine provides a constant thrust of 15.0 N. The target is approaching at a speed of 15 m/s. At what horizontal dista nce between the target a nd the rocket should you launch
Physics
1 answer:
fredd [130]3 years ago
4 0

Sum the forces in the y (upward) direction

\sum F_y = ma

F_t +F_w = ma

15N - (0.800kg)(9.81m/s^2) = 0.800kg * a

a = 8.94 m/s^2

Applying the kinematic equations of linear motion we have that the displacement as a function of the initial speed, acceleration and time is

s = v_0t + \frac{1}{2} at^2

30 = 0 +\frac{1}{2} (8.94) t^2

t = 2.59 s

Again through the kinematic equation of linear motion that describes velocity as the change of displacement in a given time, we have to

v = \frac{d}{t} \rightarrow d = vt

d = (15m/s)(2.59s)

d = 38.85m

Therefore the horizontal distance between the target and the rocket should be 38.83m

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Answer:

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The total work done on it during the compression is 3 J

Explanation:

Given;

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