Tuesday
March 28, 2017

Post a New Question

Posted by on .

The acceleration of an object g = -9.8 m/s^2. From the top of a building of 245 m., an object is thrown upwards at an initial velocity of 12m/s

(a) What is the function for the velocity of the object?

I put vf^2 = Vi^2 + 2a(yf-yi)
vf^2 = 216.09 + 19.6 (yf - 245)

(b) Determine the function for the position of the object relative to the ground.

I don't know what to put for this one

(c) How long will it take for the object to reach maximum height?

I know that for maximum height velocity = 0 with the formula: Y = Yo + Vot+ 1/2at^2, but the problem is that when I try to isolate t, I still have the t from Vot which I don't know where to put, so I'm left with t = squ(2(Yo + Vot - Y))/a

(d) What is the velocity of the object once it reaches the ground?

For this one I'll use the same formula as for (a) but will my initial height be the maximum height reached from (c)?

  • Math - ,

    my equation would be
    s(t) = -4.9t^2 + 12t + 245

    v(t) = -9.8t + 12

    at the maximum height, v = 0
    -9.8t + 12 = 0
    t = 12/9.8 = 1.22 seconds

    when it hits the ground, s(t) = 0
    -4.9t^2 + 12t + 245 = 0
    or
    4.9t^2 - 12t - 245 = 0

    using the formula ...
    t = 8.4 or a negative

    so at 8.4 seconds, v(8.4) = -9.8(8.4) + 12 = -70.32 m/sec

  • Math - ,

    so for (a) you used Y = Yo + Vot+ 1/2at^2 so when you derive it, you get time for the other ones, got it. THanks a lot Reiny

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question