A firecracker tossed into the air at 60^{0} 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?

Guest Sep 10, 2020

#1**+1 **

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 15^{2 }= (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

ElectricPavlov Sep 10, 2020