What was his initial speed?

Question

A man jumped horizontally from a window and reached the ground 0.80 s later. If he landed 3.6 m from the base of the building, what was his initial speed? How high was the window above the ground?

 

Answer 1

Δx = v₀t

Initial speed: v₀= Δx/t = 3.6/0.8 = 4.5 m/s

 

Initial height: Δy = ½gt² = ½(9.8)(0.8)²=3.1 m

Answer 2

Because gravity is directed downward it doesn't do any work in horizontal direction. Thus, your initial speed in horizontal direction remains unaffected. That's why you can assume the motion with constant speed, and simply apply formulas for the uniform motion (with constant horizontal component). This is one of most important aspects of projectile motion (if we ignore air friction which will affect the horizontal componenet). For distances and heights in your problem the air friction is not an issue and all formulas shown by colleaguea above, are applicable. There is  no need to repeat the same calculations because everything is correctly done: v₀ = 4.5m/s, height = 3.136 m.

Answer 3

 

Hey Sun -- here's the play-by-play: Vx constant 3.6m / 0.8s ... scale up 25% ==> 4.5m/s

dV of 10 runs 0.8s ... Vdn lands 8m/s ... Vave dn 4m/s ... height 4m/s(0.8s) ==> 3.2m high

Answer 4

This takes a little longer to explain, instead of looking up a strange formula on a

formula sheet or even worst trying to memorize the formulas - but I think it is easier.

 

You can find this answer using the same smarts you use to figure out how much money

you need to buy candy bars every day if candy bars are $10.00 (or $9.81) and that

you use to figure out how long it will take to go 420 miles to grandma's at 60 mph.

 

You can use the formulas or think of it this way. 

 

A - One important thing to remember with falling object  or  X & Y problems

     is that the X and Y are independent. Always!

     We CAN  Solve the Y or falling problem and the X or horizontal separately.

 

B - The second important thing to remember with falling problems  or  X & Y  problems

     is that if you know the time - you know everything, or can find it easily.

     If you don't know the time in a falling problem - find it first.  It makes the rest easier.

 

C- We know the time (0.8 s) so we can solve the whole up & down problem easily.

 

1- What is the starting velocity?  Zero -- 0

2- After 0.8 seconds falling (when he hit the ground) how fast was he going?

    0.8 sec. X  10 meter per sec. increase every sec. = 0.8 s X 10 m/s2 = 8 m/s

         (if you want to make the problem harder and be a minuscule

          2.5% more accurate you can use 9.81 m/s2 instead of 10 m/s2)

 

3- Okay now you know the starting speed and the ending speed;

    what was the average speed?   (0 + 8) ÷ 2 = 4  so an average of 4 m/s

 

4- Good! Now, if something goes an average of 4 m/s for 0.8 s, how far does it go?

    4 m/s for 0.8 s = 4 m/s X 0.8 s = 3.2 m => 3 m

 

6) So he fell 3 m therefor the window was 3 m above the ground.

          ( Using 9.81 instead of 10 would give 3.14

            - but with sig figs the correct answer is still 3 )

 

C - Lets try the horizontal (X) problem now.

 

D - One important thing to remember with falling object  or  X & Y problems

     is that there are no forces in the horizontal (X) direction for free fall.

     Therefor the X velocity is constant and does not change - Ever

1- What is the starting horizontal velocity?  Unknown

2- What is the ending horizontal velocity?  Unknown

3- But that is what we are looking for and the are the Same!

 

2- How long did he travel in the horizontal directions? 0.8 s The same as falling!

    (You hit the ground t the same time falling and moving horizontally - right?)

3- How far did he go horizontally? - 3.6 m

 

4-How fast was he going?  s= d / t = 3.6 m ÷ 0.8 m = 45 m/s

   (you were asked how fast (Speed) Not the velocity.

 

C - You can do either the falling, vertical (Y) or  the horizontal (X) problem first.