Projectile motion of shotputter

[steve snash]

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?

 

[WiFO215]

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]

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

 

[rl.bhat]

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]

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]

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]

alright, cheers