Projectile motion of shotputter

In summary, the question asks to calculate the horizontal distance of a shot put thrown at an initial speed of 15.5m/s and 34.0 degree angle to the horizontal, with a starting height of 2.2m. The solution involves using the equations of motion for constant acceleration, and solving for time and horizontal distance using the given variables.
  • #1

Homework Statement


A shotputter throws a shot with an initial speed of 15.5m/s at a 34.0 degree angle to the horizontal. Calculate the horizontal distance given that the shot leaves the athletes hand 2.2m above the ground?


Homework Equations


Distance=((initial velocity)^2 x sin (2xdegrees))/gravity(9.8)

Although this is the level range formula used when height final = height of initial. How do you change this formula to accommodate the fact that the shot lands 2.2m lower than it started?

The Attempt at a Solution


d=15.5^2 x sin 68/ 9.8 =29.09m, I know this must be wrong but i have read through the textbook and it does not help me at all in finding how to work this out. I shows what to do if the degrees is not given but with the degrees what equation can i use?
 
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  • #2
There isn't any point by-hearting formulae such as the one above. Use the equations of motion for constant acceleration,
v = u + at
s = ut + at2/2
v2 = u2 + 2as

where u and v denote initial and final velocities, a denotes acceleration, and s denotes displacement.
 
  • #3
Yeah but none of them incorporate the angle of release given? I am guessing that's an important variable, so shouldn't it be used
 
  • #4
Initial height yo of the shot put is given.At the end it reaches the ground. So the final height y = 0
So
y = yo + Vo*sinθ*t - 1/2*g*t^2.
Solve for t. Then horizontal distance
x = Vo*cosθ*t.
Now solve the problem.
 
Last edited:
  • #5
Your equations must be wrong
0=2.2+15.5*sin 34*t-1/2*9.8*t^2
t=-0.225
x=15.5*cos 34*-0.225
x=-2.8912
this answer does not make sense
 
  • #6
Negative time is not possible. While solving the quadratic equation, you will get two values of t. You have to select the positive value of t.
 
Last edited:
  • #7
alright, cheers
 

1. What is projectile motion?

Projectile motion is the motion of an object through the air, where the only force acting on the object is gravity. It follows a curved path known as a parabola.

2. How does projectile motion apply to shotputters?

Shotputters use projectile motion when throwing the shotput, as the shotput follows a parabolic path through the air due to the force of gravity.

3. What factors affect the projectile motion of a shotputter?

The factors that affect the projectile motion of a shotputter include the angle and speed at which the shotput is thrown, as well as air resistance and the force of gravity.

4. How is the trajectory of a shotput calculated?

The trajectory of a shotput is calculated using the principles of projectile motion, taking into account the initial speed and angle of the throw, as well as other factors such as air resistance and gravity.

5. How can the projectile motion of a shotputter be improved?

The projectile motion of a shotputter can be improved by adjusting the angle and speed of the throw, as well as working on techniques such as proper release and follow-through in order to maximize distance and accuracy. Additionally, a shotputter can also work on improving their strength and power through training and conditioning programs.

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