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A diamond ring is thrown from the roof of a building 12.0m above ground. You may ignore air resistance. For the motion from her hand to the ground, what are the magnitude and direction of
a) the average velocity of the ring
b) How many seconds after being thrown does it strike the ground?
c) What is the speed of the ring before it strikes the ground?

2007-03-14 17:31:01 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

a)
v square= u square + 2 as
by this, u=0m/s, a=10, s=12
v is 240^(.5)
average velocity is (v+u)/2
b)
using a=(v-u)/t, u can find t
c)
speed of the ring is just 240^(.5)

2007-03-14 17:46:55 · answer #1 · answered by logic 3 · 0 0

Well, I'll assume the ring was actually dropped off the roof of the building. With that assumption, the initial velocity in both the x and y directions is zero.

Now remember your kinematic equations:
∆x = ½*a*t² + v*t
Oh, well in this case we're using y instead of x:
∆y = ½*a*t² + v*t
Where t is the time of flight, a is the acceleration (in this case -9.80 m/s² -- the free fall acceleration due to gravity), and v is the initial velocity in the y direction (in this case zero).

Naturally, ∆y, the distance the ring falls, is equal to final height (0 m) minus the initial height (12 m), thus -12 [m].

Plug these values into the above kinematic equation and solve for the time of flight (t). This answers part B.

Now, there's two ways to solve for parts A and C. You'll want to solve for part C first, since you'll need that to calculate part A.

The first method uses kinematics, where:
(v_final)² = (v_initial)² + 2*a*∆y, where again your initial y velocity is zero, your acceleration is -9.80 [m/s²] and your ∆y is again -12 [m].

The second method uses energy. Now, if you haven't seen this yet, then don't worry about it and just stick to the first method.
Assuming v_initial equals zero, your initial_energy is all potential, and is equal to m*g*h, where m is the mass of the ring, g is the free fall acceleration due to gravity (9.80 [m/s²]), and h is your height (12 [m]). Likewise, your final_energy is entirely kinetic, and equal to ½*m*v², where m is the mass of the ring, and v is your final velocity.
By the Conservation of Energy, your final energy equals your initial energy, therefore by the transitive property:
m*g*h=½*m*v²
We can cancel out the masses to get:
g*h=½*v²
And solve accordingly for v.

Now you have both your initial and final velocities, and so you can calculate your average by finding the mean: ½(v_initial + v_final).

---
Now, all of this was assuming the ring was dropped instead of thrown, and therefore did not have any initial velocities. An initial y velocity would change everything, but you can solve for everything by including it in each respective equation (and adding in the kinetic portion of the initial energy in the second method to find the final velocity). If the ring had an initial x velocity, that would not change the time of flight, since horizontal motion is independent of vertical motion in free fall, however it would add another component to the initial and final velocities, thus changing parts A and C.

Good luck! I hope this helps!

2007-03-15 00:46:48 · answer #2 · answered by Brian 3 · 0 0

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