Physics Kinematics Homework Help ?

[jayjay55531]

Physics Kinematics Homework Help!?

"While standing on a bridge 18.3 m above the ground, you drop a stone from rest. When the stone has fallen 2.50 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction."

After finding the velocity of the first stone (AFTER falling 2.50 m) to be -7m/s, I tried to solve for initial velocity of the second stone by setting the times equal to each other (displacement for the first stone being 15.8, and for the second stone 18.3).

I used the equation x = v0*t + (1/2)*a*t2
and of course I got an ugly quadratic-equationey mess trying to get t on one side. There has to be an easier way to do this...

 

[Yosty22]

I would suggest solving for t for the first stone since you know it's initial velocity is 0 and its acceleration is -g. Whatever you get for that time, plug it into the equation for the second stone. Then, solve the second equation for the initial velocity.

 

[coconut62]

Hi, I tried to do this question as an exercise. Is the answer 2.06 m/s?

First, I find the time taken for the first stone to fall down 2.5m.

2.5 = (0)t + 1/2 (9.8) t^2
t= 0.71s

Then I find the total time for the stone to reach the ground.

18.3 = (0)t + 1/2 (9.8) t^2
t= 1.93s

The time left for the second stone to catch up is 1.93 - 0.71 = 1.22s

By using s = ut + 1/2 at^2,

18.3 = u(1.22) + 1/2 (9.8) (1.22)^2

u= 2.06m/s.

Correct?

 

[berkeman]

Hi, I tried to do this question as an exercise. Is the answer 2.06 m/s?

First, I find the time taken for the first stone to fall down 2.5m.

2.5 = (0)t + 1/2 (9.8) t^2
t= 0.71s

Then I find the total time for the stone to reach the ground.

18.3 = (0)t + 1/2 (9.8) t^2
t= 1.93s

The time left for the second stone to catch up is 1.93 - 0.71 = 1.22s

By using s = ut + 1/2 at^2,

18.3 = u(1.22) + 1/2 (9.8) (1.22)^2

u= 2.06m/s.

Correct?

Please note that we do not do other students' work for them here at the PF. Normally your post would be deleted for showing the solution, but since it has been 2 weeks since the original post (OP), I'll leave it visible.

 

[Nickie]

I would suggest using the kinematic equation v^2 = v0^2 + 2ax to solve for the initial velocity of the second stone. This equation relates the final velocity (v) to the initial velocity (v0), acceleration (a), and displacement (x). In this case, the final velocity is 0 m/s since both stones will hit the ground at the same instant. The acceleration is -9.8 m/s^2 due to gravity, and the displacement for the second stone is 18.3 m. By plugging in these values, we can solve for the initial velocity of the second stone, which turns out to be approximately -13.6 m/s. This means that the second stone must be thrown with an initial velocity of -13.6 m/s in order to hit the ground at the same time as the first stone.