A projectile is fired straight upward with an initial velocity of 100 m/s from the top of a building 20 m high and falls to the ground at the base of the building. Find (a) its maximum height above ground; (b) when it passes the top of building; (c) its total time in air.

Question

anonymous · Answer

Well, do you have any clue where to start?

anonymous · Answer

Calculus physics or algebra physics?  I would start here:
$$x = x_0 + v_0 t + 1/2at^2$$

anonymous · Answer

its differential equations and no not sure where to start

anonymous · Answer

Ok, well that makes it a little easier. The above equation describes relative position in terms of time. If you differentiate it, you have an equation for velocity. Where the velocity is 0, you are at maximum height. Eg. where d/dt of the above equation is 0 you are at max height. Get the time there and see what the height is.

anonymous · Answer

sorry i haven't been in school in 20 years and I just dont know how to set it up

anonymous · Answer

Well, here we know the initial v. We will just assume the building height to be 0. Then you have 
$$0 + 100t -0.5(9.8)t^2$$
Here, 9.8 is the gravitational constant here on earth.
Take the derivative of that, and what have you got?

anonymous · Answer

-9.8t+100? how did you get 9.8 and -.05

anonymous · Answer

You can lookup the general equation for parabolic motion. The 1/2 is in my first reply. The negative comes from the fact that gravity is IMPEDING you. It's acting in the direction OPPOSITE of the motion of the object. The 9.8 is simply the acceleration due to gravity on earth.

anonymous · Answer

Now your differentiation is correct, so now the equation is one of velocity, not distance. If you set that equation equal to 0, you will have a "t", where the velocity is changing from positive to negative. IE: top of it's motion.