Projectile motion (hammer throw)

  • Thread starter bigsaucy
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  • #1
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Hello all, just a few questions i'd like to clarify. thanks in advance

2.) In the mens hammer throw field event, athletes compete to throw a hammer as far as possible. A Hammer consists of a ball of mass 7.257kg attached to a cable of length 1.215 meters. Atheletes typicall spin the hammer 4 times before releasing. The world record for a hammer throw is 86.74 meters.

a) Assuming that the hammer is thrown at an angle of 45 degrees to the horizontal and neglecting air resistance, calculate the s peed of the ball when released for the world record throw

My solution to part a is as follows:

Vi = Initial velocity
Vf = Final velocity

The ball is launched at an angle of 45 degrees, therefore the y-component of the initial velocity is given by Vi sin (45) = y-component of initial velocity.

Since the y-component has a downward acceleration of -9.8m and it travels 86.74 meters and the final velocity is 0 m/s

then using the equation Vf^2 = Vi^2 + 2ad we get

(0)^2 = (Vi sin 45)^2 + 2(-9.8)(86.74)
1700.104 = (Vi sin 45)^2
Vi sin 45 = 41.2
Vi = 58.31m/s

Is this correct? Thanks.
 

Answers and Replies

  • #2
kuruman
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It is not correct. The final velocity is not zero. The final velocity is what it is just before the hammer hits the ground.
 
  • #3
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could you point me in the right direction? i have absolutely no idea how to solve it :(
 
  • #5
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i've read through the webpage that you linked me.

is it right to assume that the acceleration in the x-direction is 0m/s^2 since we are neglecting air resistance? also; if the acceleration in the x-direction is 0 m/s^2 how on Earth is it possible to find the initial velocity?
 
  • #6
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KURUMAN! I think i worked it out! the velocity in the y direction is 0 at the midpoint of the flight therefore if we divide the displacement by 2 and use it in my initial solution i would get the right answer!!

IS THAT CORRECT?!
 
  • #7
kuruman
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KURUMAN! I think i worked it out! the velocity in the y direction is 0 at the midpoint of the flight therefore if we divide the displacement by 2 and use it in my initial solution i would get the right answer!!

IS THAT CORRECT?!
Some things you said are correct, some are not. It is correct that the velocity in the y direction is zero at the midpoint and that the acceleration in the x direction is zero. However, if you want to use the equation vf2=vi2+2ad, you need to understand that "d" in this equation is the vertical displacement not the given 86.74 meters which is the horizontal displacement.

Suppose you tried something different. Let tf be the time the hammer stays in the air. Can you write two kinematic equations involving tf that give the horizontal position (86.74 m) and the vertical position (0 m) of the hammer at time tf?
 
  • #8
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haha i was so sure i was correct before; i used your method and got 20.60 m/s for the initial velocity in the x-direction, can you confirm?
 
  • #9
kuruman
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Incorrect. If you show what you did and exactly how, I might be able to find your mistake.
 

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