Is this even possible? Please answer!?

Question

So I've been stuck on this problem for some time. It seems my physics class will be the end of me. Please help me understand this problem.
A stone falls FROM REST off a cliff and then a second stone is thrown with initial speed of 74.48 m/s after 3.8 secs the first one. Acc due to grav is 9.8. both hit the ground at the same time. (HOW IS THIS POSSIBLE!!!) so the q asks that how long does the first stone take to hit the ground and how high the cliff is?
Please help me!

the stone move from rest

Answer 1

First we start with what we know.

We have two stones. So we have two equations.

We'll use this equation.

x= (Vo)t + 1/2at^2
Let x = distance down <---(don't know yet)
Vo = initial velocity down (at the start) <---(First stone "0meters/sec" and second stone "-74.48meters/sec")
a = acceleration (gravity = -9.8)
t = time (seconds) <---(we don't know yet)

So, first stone

x = (0meters/sec down)(t) + 1/2(-9.8)(t)^2
x= -4.9t^2

now second stone now remember the time is 3.8 seconds after so the stone has 3.8 seconds less than the first stone to cover the same distance

x = (-74.48meters/sec)(t - 3.8) + 1/2(-9.8)(t - 3.8)^2
x= -74.48t + 283 + (-4.9)(t^2 - 7.6t +14.44)
x= -74.48t + 283 - 4.9t^2 +37.24t -70.756
x= -4.9t^2 -37.06t +212.268

so we can put these two equations together to find out the time. since the distance that they fall are the same thing!

so,

-4.9t^2 = -4.9t^2 - 37.06t +212.268
0 = -37.06t +212.268
37.06t = 212.268
t= 5.73seconds <----(answer)

OMG... my brain hurts. ok, now that's how long it took the first stone to hit the ground. If my math is right. I hope it is.

to find the height of the cliff just substitute in the

x = -4.9t^2

so,

x = -4.9(5.73)^2
x = 160.75 meters down <---(answer)

Hope all that made sense. :D

Answer 2

ok . it is possible as a time will arrive when the second stone surpasses the first one(assume if there was no ground , stones would have travelled on and on , but because ground was there at the required height , both stones reached ground at same instant) and the question asks us that time and height when it happens.

for first stone: t is time taken to reach ground
initial vel.= u = 0
final vel.=v= 0+9.8t = 9.8t
displacement= x = 4.9 t^2

for second stone:
u=74.48
v=74.48 +9.8(t-3.8)
x= 74.48(t-3.8) + 4.9(t-3.8)^2

equating both x , we get t=5.7 sec
putting it in one x equation , we get x=159.2 m

So, height of cliff is 159.2m

Answer 3

It is quite possible, obviously if the rocks were dropped together they'd hit at the same time. If you delay dropping the 2nd rock you need to give it an initial velocity so that it can "catch up" to the 1st rock, which didn't have initial velocity.

I would probably start by wanting to know how high the cliff is. Since the same cliff was used,
y = v1(t2) + 1/2 g (t2)^2
and
y = 1/2 g (t1)^2
and the two y's are equal, thus
74.47(t2) + 1/2g*(t2)^2 = 1/2g (t1)^2
you can safely say that t2 = t1+3.8
now you have an equation solely in terms of t2, so just solve it. Note that in this example, g = + 9.8. If you wanted to use g = -9.8 you would need to say the initial velocity of the 2nd rock was -74.48.

Then plug that time t2 back in to the 1st equation for y.

Answer 4

The height of the cliff can be found by using this equation:

Δy = (Initial Velocity x t) + (g x t)

Use the information from the second stone and make the final velocity 0.

Answer 5

Stones at rest do not fall, so only one hits the ground. If we're talking meteorites, that's different.