# Physics help...Motion in 2 dimensions...(Uniform Circular Motion/Projectile Motion) Combination Problems?

Hi.

So I was having a difficult time solving problems with problems regarding Uniform Cirular motion and projectile motion...

Could someone please explain the problem below step by step. So that I can clearly understand what is going on...(I tried out the question but I wasn't sure if i got the right answer- The answer I got was 1.61 m.)

Could someone please tell me if this is the right answer or not??? THANKS!

David puts a stone in his 0.52 m. long slingshot and spins it around his head with a frequency of 3.7 Hz. He releases the stone 1.2 m. above the ground and at an angle of 38 degrees above the horizontal, and the stone strikes Goliath in the shoulder, 2.4 m. above the ground. How far was Goliath from David? (There are 2 possible answers.)

Relevance

V(0)=rw ,w=2*pi*f, w=23.248,v(0)=0.52*23.248=12.089

so now we are dealing a projectile motion and we need too write protection path equation.

which will reads as: y=-1/2*g*(x/Ettacos(teta ) )^2 + v(0)*sin(tata)*(x/v*cos(tata))+y(0)

putting data as:

y(0)=1.2 ,tata=38 degrees,y=2.4,v(0)=0.52 you will find x,which is your question.

• First of all lets find tip speed (tangential velocity)

3.7 Hz = 3.7 cycles per second (cps)

3.7 cps x (2 x pi) x 0.52 = 12.09 m/s (tangential velocity)

12.09sin(38) = initial vertical velocity = 7.4426 m/s

lets find max height:

v^2 = zero at highest point = u^2 - 2gh

7.4426^2 / (2g) = h = 2.826 m

2.826 + 1.2 = 4.026 m ( max h)

The horizontal velocity = 12.09cos(38) = 9.527 m/s

Time to max h = (4.026-1.2) / 4.9 = t^2 = 0.5767, sq-rt = t = 0.759 secs

Time for rock to fall to reach 2.4 m = 4.026 - 2.4 = 1.626 m

1.626 m = 1/2at^2 since the rock falls from rest at max h

1.626/4.9 = t^2 = 0.3318, sq-rt = t = 0.576 secs

max time = 0.576 + 0.759 = 1.335 secs

1.335 x horizontal velocity ( 9.527) = 12.72 m

The above is what i think the answer is.

Now lets look at the answer you arrived at?

You got 1.61 m

Ok, i'll work it out:

v^2 = u^2 - 2gh

u^2 vertical = 7.4426 m/s

7.4426^2 - 2gh = v^2

2gh = 2 x 9.8 x ( 2.4 - 1.2 ) = 23.52

55.39 - 23.52 = v^2 = 31.87, sq-rt = 5.645 m/s (vertical velocity at impact )

t = (2.4 - 1.2) / ((7.4426 + 5.645)/2) = 0.1833 secs

0.1833 x horizontal velocity (9.527) = 1.746 m

I can't see any one in the right mind doing this at such short distance? so i think the answer will be 12.72 m.

I hope this was done step by step for you to understand.