How high is the bridge above the water?

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To solve the physics problems presented, the first question requires converting 120 km/h to m/s and using the formula for acceleration to find the time taken for a car to reach that speed. The second question involves a rock dropped from a bridge, where it takes 2.4 seconds to hit the water, requiring calculations for its final speed and the height of the bridge. The average speed can be used to determine the height, utilizing the time and velocity. Hints suggest using basic equations of motion, specifically for uniformly accelerated motion, to derive the necessary values. Understanding these concepts is essential for solving the problems effectively.
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No 16 page 96: How long does it take for a car accelerating at 3.5 m/s^2 to go from rest to 120 km/h . Hint: first, change the velocity from km/h to m/s.





No. 17 pag. 105: A rock is dropped from a bridge to the water below. It takes 2.4s to hit the water. find:



a) the speed in m/s of the rock when it hits the water ( assume g,the acceleration and v , positive going down)



b) how high is the bridge above the water??


How would i solve this?Thanks in advanced
 
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For the second question (2b, more specifically), you can use the average speed. Now ask yourself, what do you need to calculate the average speed?

I hope you do understand 2a.

For the first question. You have the amount of speed increasement per second (definition of acceleration, roughly), and you have the final speed.

If you need any more hints (it wasn't quite clear what you did and did not understand), reply.
 
well for the second equation i don't know wht equation to use i have only time and acceleration of gravity which is 9.8
 
i just need the equation
 
i aleady solved the first one and got T=9.6s
 
to calculate aveg speed i need vf-vi/t
 
but i don't have vf or vi
 
ahm, average spped would be (vf-vi)/2. 2a is quite easy, you have the amount of speed added per second, you know the number of seconds, so...
2b, you've got the average speed, you have the time. I remember some equation (;)) to calculate distance from speed and time. Try to think for a while and work it out.
Oh, and please use the edit button.
 
V = at
s = (at^2)/2 for uniformly accelerated motion where
v = elocity at time t, a = constant acceleration and s = displacement in time t
 
  • #10
so if the rock is dropped Vi=0 vf=9.8*2.4=23.52m/s?? what about 2b
 
  • #11
You got that right. As i said, you can use the average speed for 2b. Then you have got a time and a velocity. Now, work that out on your own. If you are doing acceleration now, I am sure you have worked with constant velocity before.
 
  • #12
no i haven't before this chapter was vectors and first chapter was conversions just give me the equation
 
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  • #13
In that case your book is crap, but i guess that is not the issue here. Still, I will not give you the equation. I will give you hints.

velocity is defined as m/s, or distance/time. Now get some math working on that one (5=10/2 might help you).
 
  • #14
I need to print out a table with all the conversion factors
 
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