Calculating the Height of a Tower Using Projectile Motion

[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

 

[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.

 

[Kp0684]

hey thanks so much for helping me out...for the answer i got h= 30.625m...iam not sure if that's 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...

Thank you for your help...

Krishna Patel

 

[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.

 

[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...