2D motion, asking for distant land,

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To hit a horizontally moving target 29 meters above the launch point, a model rocket weighing 700 grams must be launched with a thrust of 15.0 N. The acceleration of the rocket is calculated using the formula a = (F - mg)/m, resulting in a specific time to reach the target height. The time to reach 29 meters is derived from t = sqrt(2s/a), allowing for the determination of the horizontal distance the target travels during that time. With the target approaching at 15 m/s, the distance it moves while the rocket ascends is calculated to be approximately 33.79 meters. This distance indicates the optimal launch point for the rocket to successfully intercept the target.
thanchetmy
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Please help

As a science fair project, you want to launch an 700 g model rocket straight up and hit a horizontally moving target as it passes 29 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 distance between the target and the rocket should you launch?
 
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now let us calculate the time of rocket catch 29m.
acceleration: a=(F-mg)/m F=15N ,m=0.7kg, g=10m/s^2
t=sqrt(2s/a)=sqrt[2sm/(F-mg)] s=29m
ok, we got the time, then we calculate the diatance of the target moved in this time.
d=vt=v*sqrt[2sm/(F-mg)] v=15m/s
d=15*sqrt[2*29*0.7/(15-7)]m=33.79m
 

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