Find the time the ball is in motion

[jbutl3r25]

help please!

1. A ball is thrown horizontally from the top of a building 22.8 m. high. The ball strikes the ground at a point 52.1 m. from the base of the building.
a. Find the time the ball is in motion.
b. Find the initial velocity of the ball.
c. Find the x component of its velocity just before it strikes the ground.
d. Find the y component of its velocity just before it strikes the ground.



2.
vx =
dx=52.1
t

vfY
voY = 0
aY = -9.8 (gravity)
dY= -22.8
t



3. I found the time to be 2.16 s., but the rest is tricking me up. The answer to b can't be zero can it? I don't understand that part, and I'm just at a loss for how to find the x and y components. So please...can anybody give me some help here?

 

[Astronuc]

It would be helpful if one showed the calculations.

The time is found from the freefall in the vertical direction.

Neglecting air resistance, the horizontal velocity is constant, so the velocity is found by dividing the horizontal displacement by the time it takes to travel that distance.

There is a horizontal velocity (constant) and a vertical velocity (time dependent), which changes with the acceleration due to gravity.

 

[jbutl3r25]

Yeah I used d=vo(t) + 1/2at^2 to find the time.

so to find the horizontal velocity, I would use dx=vx(t), right?
and then how would i go about finding the y?

 

[Astronuc]

Yes, in the horizontal, v_x = dx/dt.

When the ball is thrown, unless stated, one may assume is it thrown horizontally. If thrown up at and angle, it will have a vertical climb (upward velocity in y) to an apex and then freefall. If thrown downard at an angle, it will have a downard initial velocity.

See this example - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra11

 

[jbutl3r25]

i understand the x now, but i still don't understand the y...so would you use the free fall formula to find it or what?

 

[jbutl3r25]

or one of the four equations for constant acceleration?