can anyone help me with my physics question?

Question

An object is thrown downward with an initial speed of 13 m/s from a height of 64 m above the ground. At the same instant, a second object is propelled vertically from ground level with a speed of 18 m/s. The acceleration of gravity is 9.8 m/s(squared). At what height above the ground will the two objects pass each other?

ive tried talking to everyone about it but nobody can help me out.

Answer 1

What we need to do is write an equation for each object, giving its height above ground as a function of time. Then, we can set the equations equal to each other and solve for the time t when they will collide. Finally, we can substitute that value of t back into one of the equations, and get the height.

Kinematic equations tell us that:
Height = Initial Height + Velocity*t + (Acceleration * t^2) /2

For the object thrown downwards:
Height = 64 -13t - 4.9t^2

For the object thrown upwards:
Height = 0 + 18t - 4.9t^2

A few important things: the ground is declared to be at zero, so object A starts 64 m above, while object b starts at 0. Second, note the use of positive and negative signs. I arbitrarily decided that upwards velocity is positive, while downwards velocity is negative. Since gravity always acts downwards, the acceleration due to gravity terms in both equations are negative.

Setting equations equal to each other:
64 -13t - 4.9t^2 = 0 + 18t - 4.9t^2
31t = 64
t = 64/31

Substitution t = 64/31 into one of the equations (doesn't matter which)
h = 18(64/31) - 4.9(64/31)^2
h = 16.28 m

Answer 2

Okay. The distance each object travels is given by the formula:

d = v(i) t + 1/2 a t^2

Initial velocity times time, plus 1/2 acceleration times time squared. That's good! We have all that information. What's more, we know the total distance the two objects collectively travel, which is 64 meters. So we end up with the following two equations:

d(1) = v(1) t + 1/2 a(1) t^2
d(2) = v(2) t + 1/2 a(2) t^2

For our first object, initial velocity is 13 m/s in the 'toward the other ball' direction, and acceleration is 9.8 m/s in that direction. For the second object, initial velocity is 18 m/s in the 'toward the other ball' direction (which is actually the opposite direction as the other one, but we're not worried about that because the movements can be considered seperately) and the acceleration is -9.8 m/s because this time gravity is pulling the ball away from the other. So lets add the two, since we know that d(1) + d(2) = 64 m.

d(1) + d(2) = t ( v(1)+v(2) ) + 1.2 t^2 ( a(1) + a(2) )

There's something really interesting about this equation. If you're paying close attention, you'll note that a(1) + a(2) is 0! This makes sense - one ball is being sped up at EXACTLY the same rate that the other is being slowed down. The gravitational pull has NO EFFECT on when these balls will meet. So, to continue simplifying...

d(1) + d(2) = 64 m = t ( 13 + 18) = 31 t
t = 64 / 31 = 2.06 seconds

So now we know how long it takes for the balls to pass. To find the distance this is at, we'll have to go back up to the slightly more complicated equation (we'll use d(2) since that's coming up from the ground):

d(2) = v(2) t + 1/2 a(2) t^2 = 18 x 2.06 + 0.5 x -9.8 x 4.26 = 16.2 m

Which is an answer that makes sense - since the top one is accelerating and the bottom decelerating, we would expect the meeting point to be much closer to the bottom one when the starting velocities are vaguely close.

Hope that helps!

Answer 3

I don't have the answer bt here 's how you can solve it

first create the equation of motion for the first object :
the only force applied during the fly is gravity
F=-m*g
and you have m*a=F (where a is the acceleration)
hence here a=-g
the acceleration is the derivative of the speed hence
v(t)=-g*t-13 (the initial speed is -13m/s minus because it goes downward)
finally integrating one more time
z(t)=64-13*t-4.9*t*t

you do the same for the second object and you find
h(t)=18*t-4.9*t*t

now the two objetc pass each other when they are at the same height at the same time
so you have to solve z(t)=h(t) or 64-13*t-4.9*t*t=18*t-4.9*t*t

64=31*t hence they cross when t=64/31~2.1 s

now you find the height by using this time in either equation :-)

Answer 4

Displacement = v(0)*t +/- 4.9t^2
Set up both separately

1.
D1 = 13t + 4.9t^2 (plus 4.9t^2, because object is thrown with gravity)

Let D1 = y


2.
D2 = 18t - 4.9t^2 (minus 4.9t^2 because object is thrown against gravity)

Let D2 = 64 - y

y = 13t + 4.9t^2

64 - y = 18t - 4.9t^2

Add the two together:

64 = 31t

t = 2.06 seconds

Now solve for "y"
y = 13t + 4.9t^2
y = 13(2.06) + 4.9(2.06)^2
y = 47.57 meters (the distance the object from above traveled)
64 - y = 64 - 47.57 = 16.43

=16.43 meters ( the distance the the object from below traveled which is also the height above the ground where they met.

Answer 5

An object thrown from a height of 64 m at 13m/s and accelerated at 9.8m/s downwards will be in the air for...

no wait, its too difficult.

Sorry buddy

Answer 6

distance(s) = initial velocity(vi) x time(t) + 1/2 acceleration x time^2
s1 + s2 = 64
s1 = 13t + 9.8t^2
s2 = 18t - 9.8t^2
Solve for t.
Substitute into either equation above and solve for s1 or s2.
s2 = distance above ground.
s1 = distance below 64m.

Answer 7

Check out http://dir.yahoo.com/Science/Physics/ to help you.