Answer :
Answer:
As a science fair project, you want to launch an 950g model rocket straight up and hit a horizontally moving target as it passes 33.0m above the launch point. The rocket engine provides a constant thrust of 20.0N . The target is approaching at a speed of 18.0m/s . At what horizontal distance between the target and the rocket should you launch?
= 43.56m
Explanation:
acceleration =
(20 - (0.95 * 9.8) )/ (0.95)
= 10.68 / 0.95
= 11.24 m/s²
we use
s = ut + (1/2) at²
Given that
s= 40
u =0
s = 0 * t + (1/2) (11.24)t²
t = √(66/1.24)
t = √5.87
t = 2.42sec
hence
Horizontal distance = 18 * 2.42
= 43.56m
The horizontal distance covered by the rocket is 43.56m
Here is the complete question:
As a science fair project, you want to launch an 950g model rocket straight up and hit a horizontally moving target as it passes 33.0m above the launch point. The rocket engine provides a constant thrust of 20.0N . The target is approaching at a speed of 18.0m/s . At what horizontal distance between the target and the rocket should you launch?
Horizontal distance:
Thrust provided by the rocket emgine is 20N and the weight of the rocket is
[tex]W = mg = 0.95\times9.8N[/tex]
The net force on the rocket is:
[tex]F=ma=20 - 0.95 \times 9.8\\\\0.95a= 10.68\\\\a= 11.24 \;m/s^2[/tex]
where a is the acceleration
Now from the second equation of motion:
s = ut + (1/2) at²
s = 0×t + (1/2) (11.24)t²
t² = (66/1.24)
t² = 5.87
t = 2.42sec
Therefore the horizontal distance:
d = velocity × time
d = 18×2.42m
d = 43.56m
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