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A firecracker tossed into the air at 600 with the horizontal at a speed of 30 m/s exploded 15 m away. How long was the firecracker in the air before it exploded? How high did it go?

 Sep 10, 2020
 #1
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x component of speed =  x cos 60 = 30      reveals  x = initial speed = 60 m/s

 

now the path the firecracker follows is given by

    x = 60 cos 60   t  = 30 t

    y = 60 sin 60  t        (this ignores the effect of gravity.....for this short period  t  .....to make much simpler calculations )

 

Using distance formula      152   = (30t)2  + (30 sqrt3 t)2

 

                                                  15^2 = 900t^2 + 2700 t^2

                                                    t = 1/4 sec                  in 1/4 second the firecracker will get this high:    60 sin 60  (1/4) = ~ 13 meters

 Sep 10, 2020
edited by ElectricPavlov  Sep 10, 2020
 #2
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Extra! Extra! Mathematician tosses firecracker... Projectile Equations explode in his face...

 Sep 10, 2020
edited by Guest  Sep 10, 2020

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