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Physics motion problem

  • Thread starter 5.98e24
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  • #1
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Homework Statement


A monkey is standing on the top of a cliff 50 metres high and drops a coconut. His friend is standing at the base of the cliff and throws a coconut upward with an initial velocity of 15m/s. At what distance and at what time will the two coconuts collide?

Homework Equations


v2 = u2 + 2as

The Attempt at a Solution


Monkey throwing down:
v2 = u2 + 2as
v2 = 2(9.81m/s2)(50m)
v2= 981.m/s
v = 31.m

a = (v2 - v1) / t
a = (31.m/s - 0m/s) / t
a = 31.m/s / t
t = 31.m/s / 9.81m/s2
t = 3.1s

Monkey throwing up:
v2 = u2 + 2as
0m/s = (15m/s)2 + 2(-9.81m/s2)s
0m/s = 225m/s - 19.6m/s2s
-225m/s = -19.6m/s2s
s = 11.5m

a = (v2 - v1) / t
-9.81m/s2 = (0m/s - 15m/s) / t
-9.81m/s2 = -15m/s / t
t = -15m/s / -9.81m/s2
t = 1.5s

I have the initial and final velocities, the distances, the acceleration, and the times for both coconuts. I'm not sure how to find when the balls actually meet and where they meet. At first, I thought you were supposed to subtract the two times but then I realized that doesn't actually give you the time when the coconuts collided. Thank you in advance for your help.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi 5.98e24! Welcome to PF! :smile:
A monkey is standing on the top of a cliff 50 metres high and drops a coconut. His friend is standing at the base of the cliff and throws a coconut upward with an initial velocity of 15m/s. At what distance and at what time will the two coconuts collide?

Homework Equations


v2 = u2 + 2as
Nooo … you need the two coconuts to have the same s at the same t, so you need a constant acceleration equation that involves s and t :wink:
 
  • #3
It looks like they collide after the coconut going up starts going down, so:

My attempt at the solution

Monkey throwing up coconut 2:
v2 = u2 + 2as
0m/s = (15m/s)2 + 2(-9.81m/s2)s
0m/s = 225m/s - 19.6m/s2s
-225m/s = -19.6m/s2s
s = 11.5m

a = (v2 - v1) / t
-9.81m/s2 = (0m/s - 15m/s) / t
-9.81m/s2 = -15m/s / t
t = -15m/s / -9.81m/s2
t = 1.5s

Monkey dropping down coconut 1:
delta s = vt+.5at^2
delta s = (0)t + -4.9t^2
t = 1.5s
delta = -4.9(1.5)^2
delta s = -11.025 m, meaning the coconuts have not collided by that point 1.5s
50 - 11.025 = 38.975 meters
location of coconut 2 is 11.5 m
distance between the two at 1.5s = 38.975 - 11.5 = 27.475 m

Then, I'll need its velocity at that point, so:
vf^2 = vi^2 + 2as
vf^2 = 0 + 2*9.8*11.025
vf = -14.7 m/s

So, setting up the problem from there:

coconut 2
vi = 0
vf= ?
t = tcollision
delta s = x
a = -9.8

coconut 1
vi = vf (see above) = -14.7 m/s
vf = ?
delta s = x +27.475
t = tcollision
a = -9.8

Combining the two:
delta s = vt+.5at^2
coconut 1
x = 0(t) + 4.9 t^2
coconut 2
x+11.025 = 14.7t + 4.9t^2
x = 14.7t + 4.9t^2 - 27.475

Combine: 4.9t^2 = 4.9t^2 + 14.7t - 27.475
Solve for t, = 1.87 seconds

.75 + 1.5 = 3.36 seconds

Plug in t(1.87) and Solve for x, = 17.1 meters
11.5 - 17.1 meters = - 6, which is impossible

I dunno; Maybe they collide at 3.1 seconds on the ground.
 

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