Saturday
March 25, 2017

Post a New Question

Posted by on .

If an object is fired upward from the top of a tower at a velocity of 80 feet per second, the tower is 200 ft high, the formula is
h(t)=-16t^2+80t+200, t is time h is height, how long after it is fired does the object reach the max height? It is like a quadratic equation

  • MAth - ,

    If an object is fired upward from the top of a tower at a velocity of 80 feet per second, the tower is 200 ft high, the formula is
    h(t)=-16t^2+80t+200, t is time h is height, how long after it is fired does the object reach the max height? It is like a quadratic equation

    The classic equation for height reached by a body projected upward is h = Vot - gt^2/2 or Vot - 16t^2 where Vo = the initial upward velocity, t = the time of flight and g = the acceleration due to gravity.

    Your equation of height reached is given by h = 200 + 80t - 16t^2.

    The equation of motion from which you can obtain the time of flight is Vf = Vo - gt where Vf = the final velocity. Since the final final velocity is zero, we have 0 = 80 - 32t making t = 2.5 sec.

    After 2.5 sec., it will have reached its maximum height of 200 + 80(2.5) - 16(2.5)^2 = 200 + 200 - 100 = 300 ft.

  • math - ,

    i don't know

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question