Kinematics: Ball Thrown Upward from Building

In summary, the conversation discusses the kinematics of a ball being thrown from the top of a 100m building and falling back to the ground. It includes equations of motion to calculate the initial speed, position and speed after 1 and 4 seconds, and the maximum height reached by the ball. The conversation also clarifies the convention for positive and negative directions and provides a suggestion to plot a graph for better understanding.
  • #1
dboy83
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0
kinematics-- ball thrown

A ball is thrown upward from the edge of a building that is 100m tall. On its way back down, the ball just misses the building, falling to the ground below. The elapsed time between the throw and the ball hitting the ground is 6.5 sec.


A) The initial speed of the ball
B) Position and speed of the ball after 1 sec and after 4 secs.
C) The maximum height reached by the ball, measured above the ground.

*Ok, I've worked a few problems where the object is dropped directly from the top of the building but none where the object is actually thrown upward!*
 
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  • #2
Well u just hav 2 use the same equations of motion

The usual convention for these 1s is up is positive and down is negative so the initial speed u is a positive num since in the up direction. The displacement s will be -100 and g is -9.81. Putting in the numbers:

s = u*t - 0.5*g*t^2
-100 = 6.5*u - 0.5*(9.81)*(6.5)^2
-100 = 6.5*u - 207.236
6.5*u = 107.236
u = 16.498

To get the speeds just use v = u - g*t
positions: s = u*t - 0.5*g*t^2

U can use the fact that v becomes 0 at the max height to find it:
v^2 = u^2 - 2*g*s
u^2 = 2*g*s
s = u^2/(2*g)
 
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  • #3
why?

blackwizard said:
Well u just hav 2 use the same equations of motion

The usual convention for these 1s is up is positive and down is negative so the initial speed u is a positive num since in the up direction. The displacement s will be -100 and g is -9.81. Putting in the numbers:

s = u*t - 0.5*g*t^2
-100 = 6.5*u - 0.5*(9.81)*(6.5)^2
-100 = 6.5*u - 207.236
6.5*u = 107.236
u = 16.498

To get the speeds just use v = u - g*t
positions: s = u*t - 0.5*g*t^2

U can use the fact that v becomes 0 at the max height to find it:
v^2 = u^2 - 2*g*s
u^2 = 2*g*s
s = u^2/(2*g)

why is s -100? I thought it should be +100 ?

d=v0*t + 0.5at^2 isn't that the correct formula?
 
  • #4
dboy83 said:
why is s -100? I thought it should be +100 ?

d=v0*t + 0.5at^2 isn't that the correct formula?
Yes. But a = is negative if up is positive. Final d (vertical displacement) has to have the opposite sign of v0 and v0 has to have the opposite sign of a since v and a are in opposite directions.

I would suggest that you plot a graph of velocity vs. time and the up direction positive. The graph is a straight line with slope -9.8 m/sec^2.

[Of course, you could also do it with the down direction being positive and the slope + g but then the initial velocity would be negative.]

AM

PS and do not double post!
 
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What is kinematics?

Kinematics is the branch of mechanics that deals with the motion of objects, without considering the forces that cause the motion.

What is a ball thrown?

A ball thrown refers to the motion of a ball as it is propelled through the air by a force, such as a person's throw or a machine.

What factors affect the motion of a ball thrown?

The motion of a ball thrown is affected by factors such as the initial velocity, the angle at which it is thrown, air resistance, and gravity.

How is the velocity of a ball thrown calculated?

The velocity of a ball thrown can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is the time.

Does the mass of the ball affect its motion when thrown?

Yes, the mass of the ball can affect its motion when thrown. A heavier ball will require more force to be propelled through the air and will experience more air resistance, resulting in a shorter distance traveled compared to a lighter ball with the same initial velocity and angle of throw.

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