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

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A stone is thrown from a 40m tower at a 60° angle, landing 100m from the base. To find the initial speed, the horizontal and vertical motions are analyzed separately, leading to two equations involving time and initial velocity. The horizontal motion equation is 100 = v_0 cos(60°) t, while the vertical motion equation is -40 = v_0 sin(60°) t - (1/2)gt^2. After solving the equations, the initial speed is calculated to be approximately 29.x m/s, with a time of flight around 6.x seconds. The discussion emphasizes the importance of separating horizontal and vertical components in projectile motion calculations.
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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:
 
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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

x-x_0=v_0(\cos{\theta})t

this is the displacemt in the vertical direction

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

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

100 = v_0 \cos\left(\frac{\pi}{3}}\right) t

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

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
 
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