Physics help please compete problem?
Stone #1 is thrown vertically upward with a velocity of 50m/s. At the same instance a second stone Is dropped from a hill top 100m from the ground. The two stones eventually pass each other.
A. Compute the time the two stones meet
B. calculate the position of stone 1 at the point where it meets stone 2
C. Calculate the speed of stone 2 at the location where it meets stone 1
Comments
Equation of motion in the vertical direction:
hf = hi + vi*t - 1/2*g*t^2
Stone 1:
hi = 0
vi = 50 m/s
hf = 50*t - 4.905*t^2
Stone 2:
hi = 100 m
vi = 0
hf = 100 - 4.905*t^2
Set equal and solve for 't'
100 - 4.905*t^2 = 50*t - 4.905*t^2
100 = 50 t
t = 2 s <------- ANS A
Just use the equation of motion for stone 1 for B
hf = 50*t - 4.905*t^2
hf = 50 * 2 - 4.905 * 4 = 80.38 m <----------- ANS B
Simply use the velocity equation for C
vf = vi - g*t
vf = 0 - 9.81 * 2 = -19.62 m/s <----------- ANS C... The negative sign just means it is going down.
(A) Use relative motion
Since relative acceleration of both the stones is 0 just use
t=d/s where t is the time taken, d is the relative displacement between the bodies and s is the relative speed. relative speed of two bodies in opposite direction is= speed of first body+speed of second body= 50+0= 50 m/s
So, t= 100/50 = 2 sec
(B) s=ut+(1/2)at^2
s for first block is distance from the ground, u is the initial speed t is the time taken which we found out above, a is the acceleration which in this case would be -g because it is in the opposite direction of motion.
s= 50(2)+(1/2)(-g)(2)^2
Taking g=10 we get s= 80 m
Hence body 1 meets body 2 at 80 m from the ground/100-80=20 m from the top
(C) use v=u+at where v= final speed (at the meeting point), u= initial speed=0 and t= time taken till the bodies meet from first question=2 sec, a= g because it is in same direction as of motion, downward
So, v=0+g(2) taking g= 10 we get v= 20 m/s
Both objects are in free fall. If you stick your reference plane to the 50m/s stone, it will just see itself moving toward the other stone at a constant 50m/s (neglecting air resistance). The rest is just standard free fall.
a = 9.8m/s*s
integrate to get
v = a*t + v0
integrate to get
d = (a/2)*t^2 + v0*t + d0