Solving Velocity Problems: Initial Velocity 80 m/s

[sb_4000]

Hi,

Im having a problem solving one part of a question..

1) A ball is thrown upward with the initial velocity of 80 m/s.
a- How long is the ball in the air? 16.33s
b- What is the greatest height reached by the ball? 326.26m
c- when is the ball 20m above the ground?
I think I am doing this wrong, but I put 20 = v_0*t, 20/V_0 = t and i get 0.25s
d- what is the velocity of the ball, when it hits the ground?
On this one I am totaly lost, but I managed to get sqrt(2g+v^2+h) which I know is wrong.

I think a and b are correct but I am not so sure about the c and d, If you can read over it and see if it right I would appreciate it..

Thank You.

 

[arildno]

Both c) and d) is wrong; how did you get a) and b)?

 

[sb_4000]

for a) v=v_0 + at, t =2(v_0-V)/g = 16.33s Thats how I got a.

for b) h = v^2/2g = (80 m/s)^2/2(9.8)= 326 m

but I don't know how to get c or d

 

[maverick280857]

Hi

If you're having trouble doing this, you might want to consider reviewing the familiar kinematic equations:

$$v = v_{0} + at$$
$$v^2 = v_{0}^2 + 2a(S-S_{0})$$
$$S = S_{0} + v_{0}t + \frac{1}{2}at^2$$

More importantly you will want to know when to apply which and/or the signs to be used. Chose a particular direction as positive and prefix the sign accordingly. For part (c), S = y = 20m, S_{0} = 0 and v_{0} = 80 m/s.

Secondly you should draw a graph for velocity vs time and 'integrate it' to get the displacement/time profile. The zero crossings and horizontal slopes (extrema points) will gave you a fair idea of the motion of the ball. After you're comfortable you can switch back to a purely algebraic approach to solve such problems though graphically they are easier to analyze.

Hope that helps...

Cheers
Vivek