Solve Stone Thrown from 40m Tower at 60° Angle: Find Speed

  • Thread starter Thread starter badman
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves a stone thrown from the top of a 40m tower at a 60-degree angle, landing 100m from the base. Participants are tasked with finding the initial speed of the stone and its speed just before impact, while noting the absence of time in the problem statement.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the separation of horizontal and vertical motion, with some suggesting that the time taken for horizontal travel is equal to that for vertical descent. The use of equations of motion is mentioned, along with questions about the initial velocity.

Discussion Status

Some participants have provided guidance on setting up equations based on horizontal and vertical displacements. There is an acknowledgment of the complexity involved in manipulating these equations, and one participant claims to have solved the problem, although no detailed solution has been shared.

Contextual Notes

The discussion reflects a lack of explicit consensus on the initial velocity and the time required for the stone's motion, with participants exploring various interpretations and approaches to the problem.

badman
Messages
57
Reaction score
0
A stone trown from the top of a 40m tower at an angle of 60 degrees above the horizontal hits the ground at the point 100m from the base of the tower.

find the speed at which the stone was thrown?

fine the speed of the stone just before it hits the ground? I am sure i can get the answer to this with the help of the first.

it doesn't give you the time. :confused:
 
Physics news on Phys.org
You'll need to calculate the time. For the first bit though, you don't need the time. Remember, you can consider horizontal and vertical motion separately. What can you say about the horizontal motion (assuming you're neglecting air resistance and stuff)?
 
well you know that the the time it takes for the stone to travel the 100m in the horizontal direction is the same amount of time that it travels in the vertical direction.
So ask yourself how long does it take the stone to travel 100m horizontally.
 
remember that there is no acceleration in the horizontal direction.
 
Grr, that's what I meant.
 
this is the displacement in the horizontal directon

[tex]x-x_0=v_0(\cos{\theta})t[/tex]

this is the displacemt in the vertical direction

[tex]y-y_0=v_0(\sin{\theta})t-\frac{1}{2}gt^2[/tex]

solve for time in the first equation and plug it into the second equation. :biggrin:
 
but what would V_0 equal too?
 
any one know ??
 
neone can give me some hints here.
 
  • #10
You should develop the system of equations

[tex]100 = v_0 \cos\left(\frac{\pi}{3}}\right) t[/tex]

[tex]-40 = v_0 \sin\left(\frac{\pi}{3}}\right) t - \frac{1}{2}gt^2[/tex]

Two equations, two unknowns.
 
  • #11
hey thanks a lot for ur time man, i already solved it. a lot of work to twist the equations around.
 
  • #12
show us your solution.
 
  • #13
I got 29.x m/s and 6.x sec
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
23
Views
4K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K