Physics Kinematics Help

[CJPowell27]

Hey guys my AP physics teacher gave us this problem to solve today with no explanation.  (first week of physics) He assumed we took regular physics so we could solve it but I didn't

could somebody help and give a little explanation 

A 4kg ball is shot horizontally at 4m/s from a height of 19m.

What is the time of flight?

What is the horizontal range?

 

Thanks in advance

[LIGISTX]

The secret here is treat the x and y components independently. Since it is shot horizontally, just break it down into Y component first. See how long it would take to fall from 0 initial velocity and 9.8m/s^2 acceleration. Once you get that time, figure out how far it will travel horizontally with an initial velocity of 4m/s for the same time you found for the fall time in the y direction

[Bene]

at least, you can try this one. 

[juretrn]

Time of flight = time of free fall from 19 m, distance travelled 4 m/s * time of flight.

[my414]

Ok, this problem is easier than you might think, Also from a student in AP physics. What I would do is find the time it takes for the ball to hit the ground ignoring horizontal motion. You could use the equation:

Y = Yo + Vo(t) + (1/2)(a)(t)^2

Vo = 0 and Yo = 19m, Y = 0m

Once you found t = { } seconds multiply the velocity by that. The weight is there to trick you.

 

NOTE: this negates air resistance if that is required

[LIGISTX]

Known: initial y velocity = 0

             y acceleration = gravity

             height = 19m

Find: Time

 

Known: initial horizontal velocity = 4

             horizontal acceleration = 0 

             time = same time found from y

 

Find: distance traveled

[xox]

Come on dude, SUVAT.

 

s=ut+at².                     Cancel out ut because initial velocity equals zero

 

19=9.81xt²                      substitute

19/9.81=t²                    re-arrange

t=1.39.                          calculate

 

There's the first one, now you do the second

 

speed=4m/s and since it takes 1.39 seconds to hit the ground, it has to be

 

speed=distance/time

And that's 4x1.39

 

So it would travel 5.56m

 

Oh yeah, as far as I am concerned, the 5kg is irrelevant

[Nineshadow]

The gravitational force pulls it down.

The force you apply to it pushes it into the front.

These two vectors merge.

Pretty simple.

[CJPowell27]

Come on dude, SUVAT.

 

s=ut+at².                     Cancel out ut because initial velocity equals zero

 

19=9.81xt²                      substitute

19/9.81=t²                    re-arrange

t=1.39.                          calculate

 

There's the first one, now you do the second

 

speed=4m/s and since it takes 1.39 seconds to hit the ground, it has to be

 

speed=distance/time

And that's 4x1.39

 

So it would travel 5.56m

 

Oh yeah, as far as I am concerned, the 5kg is irrelevant

lol we haven't learned any equation yet...he is expecting them to remember from regular physics.  Which I didn't take