# How to compute the magnitude of a initial velocity u

#### javii

1. 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.

3. 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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#### kuruman

Science Advisor
Homework Helper
Gold Member
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.

• javii

#### cnh1995

Homework Helper
Gold Member
1/2*9.81*(3.5)^2 =60 m/s^2**m**
solving x i get 50.34 meters
That's right. But you need to find the x component of the velocity.

#### javii

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.
That's right. But you need to find the x component of the velocity.
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?

#### kuruman

Science Advisor
Homework Helper
Gold Member
That is the correct answer. However it is more conventional to divide (untimes) the distance by the time to get the speed. • javii

#### javii

That is the correct answer. However it is more conventional to divide (untimes) the distance by the time to get the speed. haha, i will remember that for next time :) Thank you for your time. :)

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