How Does a Flare Fired from a Moving Truck Behave?

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SUMMARY

The discussion focuses on the physics of a flare fired vertically from a moving truck. The flare is launched at a speed of 25 m/s while the truck travels at 15 m/s. Key calculations include determining the time the flare is in the air, the horizontal distance it travels, and the maximum height it reaches. The equations of motion, particularly Y=Yo+vo(t)+1/2g*t^2, are essential for solving these problems, emphasizing the symmetrical nature of projectile motion under gravity.

PREREQUISITES
  • Understanding of basic kinematics and projectile motion
  • Familiarity with equations of motion, specifically Y=Yo+vo(t)+1/2g*t^2
  • Knowledge of gravitational acceleration (g = 10 m/s²)
  • Ability to analyze horizontal and vertical components of motion
NEXT STEPS
  • Calculate the time of flight for the flare using the equation vf=v0+at
  • Determine the horizontal displacement using the time of flight and the truck's speed
  • Explore the concept of maximum height in projectile motion
  • Review symmetrical properties of projectile motion for further insights
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of projectile motion and its applications in real-world scenarios.

lilmixbrunette1
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Problem:
A flare was fired straight up at a speed of 25 m/s from the bed of a truck traveling along a level road at a speed of 15 m/s.

a. Assuming the flare landed back in the bed of the truck, how much time was the flare in the air?

b. How far horizontally did the flare travel before landing back in the bed of the truck?

c. What maximum height did the flare reach?



Some equations that i thought might help with this problem are
deltaX=vi+deltaT*1/2adeltaT^2
but since i do not no the cange in X or the time i can not use this equation



To try and figure out this problem i made a chart showing what horizontal and vertical things that i already knew



H V
0=ax ay=10 m/s^2
15 m/s=Vxi Vyi=25 m/s
=deltaX deltaXy=
deltaT=​

thank you for your time
 
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One other eqn, Y=Yo+vo(t)+1/2g*t^2 where both Y and Yo are the same in this case, landing in the bed. Remember vo (initial y velocity and acceleration are opposite, just solve for t. The horizontal dispacement is easily computed knowing time and that ax=0. See if you can figure out the third part.

PS: most of the gravity problems are symmetrical, so time to top is just
vf=v0+at where vf=0, and the total trip up and down twice that.
 

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