projectile motion (1 Viewer)

[muttiah]

An athlete throws a shot of mass 7.26kg at an initial speed of 12m/s at an angle of 30 degrees above the horizontal. At the moment the shot leaves his hand it is 2m above the ground.

(1) the distance that he throws the shot [answe:15.6m]

(2) the velocity with which the shot hits the ground [answer:13.56m/s 39 degrees]

 

[Jono_2007]

1)
Vertically: For time in air!
t=?, u=12sin30=6.00m/s, a=-9.8m/s/s,r=-2m.
r=ut+1/2at^2
-2=6t+1/2.-9.8t^2
0=2+6t-4.9t^2
Using quadrtic formula, we get two values for t.
Taking the positive result, t=1.497s
Now: Horizontally: r=?, t=1.497s, u=12cos30=10.39m/s.
u=v average=r/t -->therfore r=u.t=10.39m/s x 1.497s=15.56m

2)Vertically: v=?, a=-9.8m/s/s, t=1.479s, r=15.56m
r=vt-1/2at^2
therfore v=r+1/2at=15.56+1/2x-9.8x1.497=8.2247m/s
horizontally: Since no forces are acting on the object horizontally, the horizontal motion is the same. Therefore u=v=10.39m/s
Now draw vectors representing these horizontal and vertical components ans use pythagorus theorem to work out the resultant.

V^2=8.2247^2+10.39^2--->therfore V=13.25m/s @ 38.4* below horizontal
angle=tan^-1(8.2247/10.39)=38.36*

 

[idling fire]

2)Vertically: v=?, a=-9.8m/s/s, t=1.479s, r=15.56m
r=vt-1/2at^2
therfore v=r+1/2at=15.56+1/2x-9.8x1.497=8.2247m/s

I'm not sure I get the reasoning behind this... The r=15.56 was for horizontal displacement, so shouldn't be involved in the vertical component...

EDIT:
For vertical velocity, using v=u+at where v=? u=12sin30=6 a=-9.8 (its the NEGATIVE direction still) t=1.497

So v=6+(-9.8)x1.497=-8.6717m/s
This negative value is good, because it shows it's going downwards.

horizontally: Since no forces are acting on the object horizontally, the horizontal motion is the same. Therefore u=v=10.39m/s
Now draw vectors representing these horizontal and vertical components ans use pythagorus theorem to work out the resultant.

EDIT:
v^2=(-8.6717)^2=10.39^2
v=13.535m/s

tan@=8.6717/10.39
@=39.8 degrees
(ignore the negative value since you only want an acute angle)

Therefore velocity at ground is 13.535m/s at 39.8 degrees below the horizontal.

 

[idling fire]

Okay, sorry I did something stupid. :/
I've edited it in in the above post.

 

[Forbidden.]

1)
Vertically: For time in air!
t=?, u=12sin30=6.00m/s, a=-9.8m/s/s,r=-2m.
r=ut+1/2at^2
-2=6t+1/2.-9.8t^2
0=2+6t-4.9t^2
Using quadrtic formula, we get two values for t.
Taking the positive result, t=1.497s
Now: Horizontally: r=?, t=1.497s, u=12cos30=10.39m/s.
u=v average=r/t -->therfore r=u.t=10.39m/s x 1.497s=15.56m

2)Vertically: v=?, a=-9.8m/s/s, t=1.479s, r=15.56m
r=vt-1/2at^2
therfore v=r+1/2at=15.56+1/2x-9.8x1.497=8.2247m/s
horizontally: Since no forces are acting on the object horizontally, the horizontal motion is the same. Therefore u=v=10.39m/s
Now draw vectors representing these horizontal and vertical components ans use pythagorus theorem to work out the resultant.

V^2=8.2247^2+10.39^2--->therfore V=13.25m/s @ 38.4* below horizontal
angle=tan^-1(8.2247/10.39)=38.36*

For the first one, I thought c in ax²+bx+c=0 was 2, instead of - 2 and I got my discriminant as less than zero.
But now I agree.

 

[Jono_2007]

I'm not sure I get the reasoning behind this... The r=15.56 was for horizontal displacement, so shouldn't be involved in the vertical component...

Yes i know i confused the question with my working, and i thought that the verticall displacement from where the object was fired from, was 15.56m, needless to say your answer is right. I re-did the question and got the same answer.
Thanks for picking that up.

JONO_