What is the horizontal distance that the firework will travel?

A firework is shot up from the ground, travels up to a height of 77 meters, and comes to stop before exploding.



When it explodes, a fragment is ejected with a velocity of 13 m/s at an angle of 45 degrees above the horizontal.



What is the horizontal distance that this fragment will travel before hitting ground?What is the horizontal distance that the firework will travel?Vertically, relative to its starting point 77m high, one wants



h = - 77m = 13/(sqrt 2) t - 1/2 g t^2, so that



- 77 = 9.1924... t - 4.905 t^2, or t^2 - 1.8741... t - 15.698... = 0.



Solving for t, t = [1.8741... + /- sqrt (3.5122... + 62.793...)] / 2 s.



Only the + sign is relevant so that t = [1.8741... + /- sqrt (66.305...] / 2, or



t = 5.0084... s.



During this time, it will have travelled a horizontal distance of



d = 13 / (sqrt(2)) x t m = 46.040... m.



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