Projectile motion (hammer throw)

[bigsaucy]

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.

 

[kuruman]

It is not correct. The final velocity is not zero. The final velocity is what it is just before the hammer hits the ground.

 

[bigsaucy]

could you point me in the right direction? i have absolutely no idea how to solve it :(

 

[kuruman]

Look at

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Find "Trajectories" under "Velocity and Acceleration" under "Mechanics".
... enjoy.

 

[bigsaucy]

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?

 

[bigsaucy]

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?!

 

[kuruman]

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 t_f be the time the hammer stays in the air. Can you write two kinematic equations involving t_f that give the horizontal position (86.74 m) and the vertical position (0 m) of the hammer at time t_f?

 

[bigsaucy]

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?

 

[kuruman]

Incorrect. If you show what you did and exactly how, I might be able to find your mistake.