Homework Help: Projectile motion of shotputter

steve snash · Mar 21, 2010

1. The problem statement, all variables and given/known data
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???

2. Relevant 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????????

3. 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 text book 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??????????????

WiFO215 · Mar 21, 2010

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.

steve snash · Mar 21, 2010

Yeah but none of them incorporate the angle of release given???? Im guessing thats an important variable, so shouldnt it be used

rl.bhat · Mar 21, 2010

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.

steve snash · Mar 21, 2010

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

rl.bhat · Mar 22, 2010

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.

steve snash · Mar 22, 2010

alright, cheers