How Does a Flare Fired from a Moving Truck Behave?

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A flare fired vertically from a moving truck at 25 m/s while the truck travels at 15 m/s will land back in the truck's bed. The time the flare is in the air can be calculated using kinematic equations, with the vertical motion being symmetrical. The horizontal distance traveled by the flare can be determined by multiplying the truck's speed by the time in the air. The maximum height reached by the flare can be found using the initial vertical velocity and gravitational acceleration. Understanding these principles allows for solving similar projectile motion problems effectively.
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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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