# General Physics 1

1. Jan 23, 2005

### Kp0684

A rock is thrown vertically upward from ground level at time t=0. At time t=1.5s it passes the top of a tall tower, and 1.0s later t reaches its maximum height. What is the height of the tower? ....... is this the equation i would use....... delta(Y)= Vo(t) + 1/2(a)(t2)...........with Vo= 0 , t=1.5, a= 9.8m/s2.....with delta(Y) in Height......= 11.025meters......need help on this one

2. Jan 23, 2005

### dextercioby

Use this equation
$$v_{fin}=v_{0}-gt$$

to find the initial velocity (which is not zero),knowing the total time of flight...
Then this
$$h=v_{0}t-\frac{1}{2}gt^{2}$$

to find the height...

Daniel.

3. Jan 23, 2005

### Kp0684

hey thanks so much for helping me out.....for the answer i got h= 30.625m......iam not sure if thats the answer you would have got.....but this is my setup..... Vo= V + gt..... i got 24.5m/s2......then i used Vo-1/2gt2......which i got it to be = 30.625m.....

Krishna Patel

4. Jan 23, 2005

### dextercioby

Do u agree that
$$h_{tower}=24.5\frac{m}{s}\cdot 1.5s-\frac{1}{2}9.8\frac{m}{s^{2}}\cdot (1.5s)^{2}$$

??

Do you understand what i did??

Daniel.

5. Jan 26, 2005

### Kp0684

yeah i see what i did wrong i plugged in 2.5s....instead of 1.5 which it took the time to pass the tower.....i thought it would be the where it reaches its max height for time......but i see how its done...thanks so much....