How to compute the magnitude of a initial velocity u

In summary, the rock is thrown horizontally from a tower point A, and hits the ground 3.5 s later at point B. A line from A to B makes an angle of 50 degrees with the horizontal. The gravity causes the rock to slow down and it lands at point B.
  • #1
javii
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Homework Statement


A rock is thrown horizontally from a tower point A, and hits the ground 3.5 s later at point B. A line from A to B makes an angle of 50 degrees with the horizontal.
Compute the magnitude of a initial velocity u of the rock.

The Attempt at a Solution


I started looking at the motion in x direction, which is
u the unknown.
Then i looked at the motion in y direction:
And here we have the gravity.
so i used the formula:
s=1/2 * g * t^2

1/2*9.81*(3.5)^2 =60 m/s^2

from trigonometry

tan(50)=60/x
solving x i get 50.34 meters

But I'm not sure if I have done it corretly.

Thank you for your help.
 
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  • #2
javii said:
1/2*9.81*(3.5)^2 =60 m/s^2
The units should be meters. Other than that, so far you are on the right track, but you have not answered the question yet. You found where the rock lands horizontally, but you are asked to find its initial speed.
 
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  • #3
javii said:
1/2*9.81*(3.5)^2 =60 m/s^2**m**
javii said:
solving x i get 50.34 meters
That's right. But you need to find the x component of the velocity.
 
  • #4
kuruman said:
The units should be meters. Other than that, so far you are on the right track, but you have not answered the question yet. You found where the rock lands horizontally, but you are asked to find its initial speed.
cnh1995 said:
That's right. But you need to find the x component of the velocity.
kuruman said:
The units should be meters. Other than that, so far you are on the right track, but you have not answered the question yet. You found where the rock lands horizontally, but you are asked to find its initial speed.
aha, so I have to times the denominator with the time (3.5):

tan(50)=60/x*3.5
Then I will get
14.38 m/s

It that true?
 
  • #5
That is the correct answer. However it is more conventional to divide (untimes) the distance by the time to get the speed. :smile:
 
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  • #6
kuruman said:
That is the correct answer. However it is more conventional to divide (untimes) the distance by the time to get the speed. :smile:
haha, i will remember that for next time :) Thank you for your time. :)
 

1. What is the formula for calculating magnitude of initial velocity?

The formula for calculating magnitude of initial velocity (u) is u = √(ux^2 + uy^2), where ux and uy are the horizontal and vertical components of initial velocity, respectively.

2. Can the magnitude of initial velocity be negative?

No, the magnitude of initial velocity is always a positive value. Negative values can represent the direction of the velocity, but the magnitude is always positive.

3. How is initial velocity related to displacement and time?

The initial velocity (u) is one of the variables in the equation for displacement (s) with respect to time (t). The equation is s = ut + 1/2at^2, where a is acceleration.

4. How can initial velocity be calculated if only the angle and speed are known?

If only the angle and speed are known, the initial velocity can be calculated using trigonometric functions. The horizontal component of the velocity (ux) can be found using ux = u * cos(θ), where u is the speed and θ is the angle. The vertical component (uy) can be found using uy = u * sin(θ).

5. What are the units of initial velocity?

The units of initial velocity depend on the system of measurement being used. In the SI system, the units of initial velocity are meters per second (m/s). In the English system, the units are feet per second (ft/s) or miles per hour (mph).

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