Calculating initial velocity for object

Click For Summary
To calculate the initial velocity of a ball thrown upward to remain airborne for 12.6 seconds, the correct acceleration due to gravity to use is -9.8 m/s², as gravity acts downward. The initial velocity calculated is 61.7 m/s, which is confirmed to be correct. When the ball hits the ground, its speed will be the same as the initial speed but negative, indicating direction. The maximum height reached by the ball is calculated to be 193 meters, which also requires confirmation. Understanding that gravity is always negative in kinematics simplifies the calculations.
babbagee
Messages
71
Reaction score
0
Calculating initial velocity for object...

I will give the problem first and then explain what i am having trouble with.

a.) With what speed must a ball be thrown directly upward so that it remains in the air for 12.6 seconds?
b.) What will be its speed when it hits the ground?
c.) How high does the ball rise?

a.) For the first part i just used the equation, v=v0 + at, and I assumed that v = -v0, so my question is do i use -9.8m/s2 or +9.8m/s2. But after thinking about it I think I should use -9.8m/s2 because when the ball reaches its maximum hight it falls back with a negative velocity, right. If someone else has another way of think about this please tell me. Oh, and I got 61.7 m/s, can some one confirm that.

b.) The velocity will be the same as the intial velocity but negative.
c.) I did this one and used -9.8m/s2, and i got 193 meters, can someone confirm that also.

I am having a hard time chosing weather to use -9.8m/s2 or +9.8m/s2. I guess what I am trying to say is what is a simple of thinking about this.

Thank you
 
Physics news on Phys.org
In kinematics questions ALWAYS take g=-9.8m/s2.Reason being that gravity always acts downwards.So if you take downwards vel. as -ve you must also take g to be -ve.
 
ok, i understand

can someone confirm my answers.

thanks
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
792
  • · Replies 3 ·
Replies
3
Views
797
  • · Replies 11 ·
Replies
11
Views
4K
Replies
11
Views
1K
Replies
2
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K