Vectors, New Student, probably very basic

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I have just started studying my first physics class this year, and I am stuck on 1 question relating to vectors. I think my problem is understanding the question, I have completed all the other questions in the section fairly easily. This is the question I am stuck on.

A stone is thrown horizontally with an initial velocity of 5ms-1. What is the magnitude and direction of its velocity 0.2s later? Take the acceleration of free fall to be 9.8ms-2 and ignore friction.

The book has the answers, but it doens't show working out. but I have not been able to figure out how to do this one. Any help? Please? Show any workings. Thanks.
 

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[tex]v_o = 5 m/s[/tex]
[tex]t = 0.2 s[/tex]
[tex]a = g = 9.82 m/s^2[/tex]

The [tex]v_0[/tex] is a vector upwards while the acceleration is a vector downwards with respect to the time.
 
Last edited:
Mattara said:
The [tex]v_0[/tex] is a vector upwards

No, it's horizontal. :smile:

The key here is to note that the vector equation

[tex]\vec v = {\vec v}_0 + \vec a t[/tex]

is really two equations, one each for the x and y components of motion:

[tex]v_x = v_{0x} + a_x t[/tex]

[tex]v_y = v_{0y} + a_y t[/tex]

The information you were given tells you the values of [itex]v_{0x}[/itex], [itex]v_{0y}[/itex], [itex]a_x[/itex], [itex]a_y[/itex] and [itex]t[/itex]. Therefore you can calculate [itex]v_x[/itex] and [itex]v_y[/itex], and from them you can calculate the magnitude and direction of [itex]\vec v[/itex].
 
The answer according to the book is "5.37ms-1, 2.14 degrees below horizontal."
 
Oh, my bad didn't read the whole question ;P
 
stevie said:
I have just started studying my first physics class this year, and I am stuck on 1 question relating to vectors. I think my problem is understanding the question, I have completed all the other questions in the section fairly easily. This is the question I am stuck on.

A stone is thrown horizontally with an initial velocity of 5ms-1. What is the magnitude and direction of its velocity 0.2s later? Take the acceleration of free fall to be 9.8ms-2 and ignore friction.

The book has the answers, but it doens't show working out. but I have not been able to figure out how to do this one. Any help? Please? Show any workings. Thanks.
jtbell has the correct equations for each of the components (x and y). Keep in mind that the horizontal component has an acceleration of zero. The vertical component has an initial velocity of zero (the stone was thrown horizontally).

jtbell's equations give you the vector in Cartesian coordinates. You want the vector in polar coordinates. Plug the x and y coordinates into the Pythagorean Theorem to get the magnitude of the vector. The tangent of the angle is equal to y/x. (Since gravity accelerates the stone downwards, your y component will be negative, hence the negative angle - i.e. 2.14 degrees below horizontal).
 
stevie said:
The answer according to the book is "5.37ms-1, 2.14 degrees below horizontal."

Either you or the book made a typo. It should be 21.4 degrees, not 2.14.