Acatapult used by medieval armies hurls a stone of mass 32.0 kg with a velocity of 50.0 m/s at a 30.0 degree angle above the horizontal. what is the horizontal distance traveled when the stone returns to its original height? ignore air resistance. check your calculator deg/rad mode.

The horizontal distance traveled when the stone returns to its original height = 220.81 m

Explanation:

Considering vertical motion of catapult:-

At maximum height,

Initial velocity, u =  50 sin30 = 25 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v = u + at

Substituting

    v = u + at

    0 = 25  - 9.81 x t

    t = 2.55 s

Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s

Considering horizontal motion of catapult:-

Initial velocity, u =  50 cos30 = 43.30 m/s

Acceleration , a = 0 m/s²

Time, t = 5.10 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

   s = 43.30 x 5.10 + 0.5 x 0 x 5.10²

  s = 220.81 m

The horizontal distance traveled when the stone returns to its original height = 220.81 m

amplitude

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