Rock tossed straight up-Velocity question

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In summary, the rock is traveling at a velocity of -21 m/s when it hits the bottom of the hole. It is in the air for 4.9 seconds.
  • #1
xgoddess210
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Homework Statement



A rock is tossed straight up with a speed of 21 m/s. When it returns, it falls into a hole 15 m deep. What is the rock's velocity as it hits the bottom of the hole? How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Homework Equations



d = vit + (0.5)at^2

The Attempt at a Solution



d= 15 m
a= 9.8
vf=
vi= -21 m/s
t=

I was able to find the time by plugging in:

15= (-21)t+4.9t^2

which was a quadratic, so it turned out to be 4.9 seconds

I'm just not sure about how to find the velocity at the bottom of the hole. I've been trying for about an hour now and this one problem is driving me mad.

I think the initial velocity would change to 0 and the acceleration due to gravity would still be the same. I tried using v=vo+at but I got a wrong answer. (V=0+9.8*4.9=48)

I'm using masteringphysics so I can try answers to see if they are correct, but I'm on my last attempt and I'm pretty desperate.

Please explain to me what I am doing wrong and help me solve this problem!
 
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  • #2
Welcome to PF.

You almost had it.

It is V = Vo - g*t

According to the statement though the initial Vo is 21 m/s UP.

So your answer will likely need to be expressed as minus, as the equation yields.

V = 21 - 9.8*(4.9)
 
  • #4
You only want to consider the downward motion of the rock starting from it's highest point (when it stops the upward motion). The only acceleration will be due to gravity. Don't forget to add the extra depth for the hole.
 
  • #5
Thank you so much! I really appreciate it!
 

Related to Rock tossed straight up-Velocity question

1. What is the initial velocity of a rock tossed straight up?

The initial velocity of a rock tossed straight up is the speed at which the rock is thrown upwards. This velocity is typically measured in meters per second (m/s) and can vary depending on the strength and angle of the throw.

2. How does the velocity of a rock change as it goes up and comes down?

As a rock is tossed straight up, its velocity will decrease until it reaches its peak height. Then, as it falls back down, its velocity will increase until it reaches the same speed as when it was initially thrown.

3. What factors can affect the velocity of a rock tossed straight up?

The velocity of a rock tossed straight up can be affected by factors such as the initial force and angle of the throw, air resistance, and gravity. These factors can either increase or decrease the velocity of the rock.

4. Can the velocity of a rock tossed straight up be negative?

Yes, the velocity of a rock tossed straight up can be negative. This would occur when the rock is falling back down towards the ground and is moving in the opposite direction of its initial throw.

5. How is the velocity of a rock tossed straight up calculated?

The velocity of a rock tossed straight up can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time. Alternatively, it can also be calculated using the equation v^2 = u^2 + 2as, where s is the displacement or change in height of the rock.

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