Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s
To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s
Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m
This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.
But anyways, good luck!
I'm not sure if this is correct but it's what I'll do
This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.
Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )
Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12
Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2
4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t
Time for Stone B is 4s
Time for Stone A is 6s
Oh my gosh ! Resisting the force of gravity always DOES involve doing work.
If no work is being done, then you're NOT resisting the force of gravity.
Example:
-- ball rolling on the floor . . . no work
-- ball rolling up a ramp . . . work being done
-- ball rolling down a ramp . . . work being done, BY gravity
Answer:
Explanation:
distance between ship A and B = 32 mile
Ship A velocity in south, dx/dt = -16 mph
Ship B is sailing toward east with speed, dy/st = 12 mph
time = 1 hour
rate of change of distance between them = ?
x be the distance travel after t time
X = 32 + x
Let distance between them be z
now, using Pythagoras theorem to calculate distance between ships after 1 hours
z² = x² + y²
z² = (32 + x)² + 12²
z² = (32 - 16)² + 12²
z = √400
z = 20 miles
now, calculation of rate of change of distnace
z² = (32 + x)² + y²
differentiating both side w.r.t. time
hence, the rate is the distance between them changing at the end of 1 hour is equal to
