TPR PHYSICS QUESTION

[pizza1994]

While in motion, a 40 kg child on The Swinging Trapeze experiences a centripetal force of 100 N, while the child's swing traces out a circle of radius 10 m. If this child drops a ball while the ride is in motion, what horizontal distance will the ball travel before it lands, given that he is 10 m above the ground?
a) 4 m
b) 6 m
c) 7 m
d) 9 m

Can someone please help me on this? I can find centripetal acceleration= 2.5 m/s^2 and v= 5 m/s
but I'm not sure how to solve this

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[tdod]

i'm assuming that the answer is c?

if so, then this is the explanation:
The ball is dropped at 10m, so first figure out how long it will take the ball to hit the ground { x = x_o + v_o(t) + (1/2)at^2 }. Then, use the centripital acceleration to find out the instantaneous velocity of the ball when released { F_c = ma_c = [mv^2 / r] } ---> { v = √ ( rF_c / m) }. You now know the time that the ball is in flight and the initial velocity. because you assume that there is no air resistance, there is no acceleration in the x direction. consequently, you can use { { x = x_o + v_o(t) + (1/2)at^2 = v_o(t) } to solve.

i hope that makes sense.

 

[Czarcasm]

i'm assuming that the answer is c?

if so, then this is the explanation:
The ball is dropped at 10m, so first figure out how long it will take the ball to hit the ground { x = x_o + v_o(t) + (1/2)at^2 }. Then, use the centripital acceleration to find out the instantaneous velocity of the ball when released { F_c = ma_c = [mv^2 / r] } ---> { v = √ ( rF_c / m) }. You now know the time that the ball is in flight and the initial velocity. because you assume that there is no air resistance, there is no acceleration in the x direction. consequently, you can use { { x = x_o + v_o(t) + (1/2)at^2 = v_o(t) } to solve.

i hope that makes sense.

Yeah, I calculated it the same way in my head: 100N = mv^2/r ; Solving for V (instantaneous velocity) you get: 5 m/s. From there, since the ball is falling (as it is released in a tangential path), you need to find out how long it would take to hit the ground. The only force acting on it at this point is gravity, which has an acceleration of ~10m/s.

The vertical component has an initial velocity of zero. We also know the height from the question, but we need to find the time. Personally, I'd calculate final vertical velocity first: vf^2 = vi^2 + 2ad = vf^2 = 2(10)(10); vf^2 = 200; vf = ~14m/s (which means total time in air is ~1.4s).

To find the range, we use the total time x initial horizontal velocity (which is constant because no force is acting on it horizontally), so: 5m/s x 1.4s = ~7 m. Choice C.

Edit: Woops, meant mv^2/r not 1/2mv^2/r.

 

[pizza1994]

why is the inital vertical velocity zero?

 

[pizza1994]

Also for the 5 m/s why is that the inital horizontal velocity? Why can't it be the vertical velocity?