- #1

missrikku

A ball is thrown down vertically with an initial speed of Vo from a height of h.

**a) What is its speed just before it strikes the ground?**

Speed is just the magnitude of instantaneous velocity, right? So I just used the equations for velocity:

Vf =?

V^2 = Vo^2 + 2a(Y-Yo)

Since I'm dealing with free fall accel, a = -9.8 m/s^2. In addition, Y-Yo = h. So,

V^2 = Vo^2 + 2(-9.8)h --> V^2 = Vo^2 -19.6h

Leaving me with:

**V = (Vo^2 -19.6h)^(1/2)**

**b) How long does the ball take to reach the ground?**

I used: Y-Yo = Vot + (1/2)at^2

and using the quadratic equation, I got:

** i used the "+" version bec t cannot be negative

**t = [-Vo + (Vo^2 - 19.6)^(1/2)]/-9.8**

**c) and d) are the same as a) and b) except this time the ball is thrown upward. I also have to figure out if the values for c and d are less than, greater than, or the same as the values in a and b.**

<b>c)</b> I used:

Y - Yo = (1/2)(Vo + V)t

h = (1/2)(Vo + V)t

**V = (2h/t) - Vo**

I think the values will be the same.

I think the values will be the same.

**d)**For this part, I used the V found in c)

V = (X1 - X)/t1 - t --> V = h/t --> t = h/V

this is the time just to fall from the maximum point the ball reaches, so the total time to fall to the ground would be:

**2t = 2(h/V)**

Because the time is doubled, I think this value will be greater than the value in b)

Because the time is doubled, I think this value will be greater than the value in b)

Were my assumptions correct?