# Homework Help: Determine the horizontal distance travelled during its motion

1. Mar 25, 2009

### Ry122

A rock is projected from the edge of the top of a building with an initial velocity of 17.1 m/s at an angle of 37degrees above the horizontal. The rock strikes the ground a horizontal distance of 59.7 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?

My attempt:
This is the method i used to solve the problem:
1. I calculated the time to for the rock to carry out it's parabolic motion. To do this i just used v=u + at and u=17.1cos37 v=0 a=-9.8
t then was equal to 1.05. I multiplied this by two since a parabola is symmetric.

i then used this time to determine the horizontal distance travelled during its parabolic motion.

Then I subtracted this distance from the complete horizontal distance travelled which
gave me the horizontal distance travelled during its non parabolic flight.

This then allowed me to calculate the time of non-parabolic flight, which I then used
to calculate the vertical distance travelled during non-parabolic flight which would be equal to the height of the building.

The final answer I got was 34.202m which is incorrect. Can someone point out where I've gone wrong?
Thanks

2. Mar 25, 2009

### rl.bhat

Re: Kinematics

u=17.1cos37 v=0 a=-9.8
This step is wrong.
What is the vertical component of initial velocity?

3. Mar 25, 2009

### Ry122

Re: Kinematics

sorry, it should be sin, that was just an error in my typing, what i have in my calculations however is correct.

4. Mar 25, 2009

### quZz

Re: Kinematics

I don't understand what are you doing in this part... are you trying to solve the equation u+at = 0? if yes then what for?

5. Mar 25, 2009

### Ry122

Re: Kinematics

im solving for t so that i can determine for what duration the rock is in the air while above the building so that i can then determine how long the rock is in the air below the building.
the time that the rock is in the air below the building for can then be used to determine how far the rock fell vertically while below the building hence determining the height of the building.

6. Mar 25, 2009

### DorianG

Re: Kinematics

Isn't the flight path after the projectile falls below its original trajectory point still parabolic?
I'm not 100% certain on that, maybe someone else can confirm...

7. Mar 25, 2009

### rl.bhat

Re: Kinematics

Through out the motion it is projectile motion.
In this problem the range of the projectile, initial velocity and the angle of projection is given. From these values you can find the time flight.
Once you know the time of flight you can find the height of the building.

8. Mar 25, 2009

### Ry122

Re: Kinematics

yeah it probably is still parabolic but thats not gonna make my method of calculation incorrect in any way.

rl.bhat i did find the time of flight to determine the height of the building, but my final answer was incorrect.

can you please work the problem out and see if you get an answer different from mine?
Thanks

9. Mar 25, 2009

### rl.bhat

Re: Kinematics

10. Mar 26, 2009

### Ry122

Re: Kinematics

Can you show me your working?

11. Mar 26, 2009

### rl.bhat

Re: Kinematics

I am using the general equation of projectile.

Y = x*Voy/Vox -1/2*g*x^2/Vox^2
In this problem you have to take Y negative, because you are measuring it in the downward direction.