Calculating the Height of a Tower Using Projectile Motion

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Homework Help Overview

The discussion revolves around calculating the height of a tower using principles of projectile motion. The scenario involves a rock thrown vertically upward, with specific time intervals provided for its motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore different equations related to projectile motion, questioning the initial velocity and the correct time intervals to use in their calculations. There is discussion about the setup of the equations and the values substituted into them.

Discussion Status

Some participants have provided equations and attempted calculations, while others are clarifying their understanding of the setup and the assumptions made. There is an ongoing exploration of the correct approach to find the height of the tower, with no explicit consensus reached.

Contextual Notes

Participants are working under the constraints of the problem as posed, including specific time intervals and the nature of the projectile motion involved. There is mention of confusion regarding the time to reach maximum height versus the time to pass the tower.

Kp0684
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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 :confused:
 
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Use this equation
[tex]v_{fin}=v_{0}-gt[/tex]

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

to find the height...

Daniel.
 
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
 
Do u agree that
[tex]h_{tower}=24.5\frac{m}{s}\cdot 1.5s-\frac{1}{2}9.8\frac{m}{s^{2}}\cdot (1.5s)^{2}[/tex]

??

Do you understand what i did??

Daniel.
 
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...
 

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