SOLUTION: A projectile is fired at an angle of 30 degrees above the horizontal from the top of a cliff 600 ft high. The initial speed of the projectile is 2000 ft/s. How far will the project

Question 1180942: A projectile is fired at an angle of 30 degrees above the horizontal from the top of a cliff 600 ft high. The initial speed of the projectile is 2000 ft/s. How far will the projectile move horizontally before it hits the level ground at the base of the cliff?
Answer by ikleyn(52855) (Show Source): You can put this solution on YOUR website!
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A projectile is fired at an angle of 30 degrees above the horizontal from the top of a cliff 600 ft high.
The initial speed of the projectile is 2000 ft/s. How far will the projectile move horizontally
before it hits the level ground at the base of the cliff?
~~~~~~~~~~~~~~~~~


Vertical component of the initial velocity is half of 2000 ft/s, or 1000 ft/s.


Therefore, the equation for the verical coordinate h(t) is

    h(t) = -16t^2 + 1000t + 600


The equation to find the time of the flight is  h(t) = 0,  or

    -16t^2 + 1000t + 600 = 0,  or

     4t^2 - 250t - 150 = 0.


Its roots are   =  =  = .



Of these two roots, only positive is interesting for us  t =  = 62.8 seconds  (rounded).



The horizontal component of the speed is   = 1732 ft/s (rouned) and is considered as a constant during the flight.


Moving with the horizontal speed of 1732 ft/s during 62.8 seconds, the projectile will get the ground at the distance of  

       62.8*1732 = 108769.6 feet from the cliff base.      ANSWER

Solved.

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