Freefall physics question, please help?

Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude v= 5.00 m/s, releases a sandbag at an instant when the balloon is a height h= 40.0 m above the ground. After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.

A: Compute the position of the sandbag at a time 0.400 s after its release.
Take the free fall acceleration to be = 9.80 m/s^2.

B: Compute the velocity of the sandbag at a time 0.400 s after its release.

C: Compute the position of the sandbag at a time 1.35 s after its release.

D: Compute the velocity of the sandbag at a time 1.35 s after its release.

E: How many seconds after its release will the bag strike the ground?

F: With what magnitude of velocity does it strike?

G: What is the greatest height above the ground that the sandbag reaches?

please explain thanks

Answer 1

Use the formulas for position and velocity.

X = Xo + Vo * t + 1/2 a * t ^ 2
V = Vo + a t

X being the final position
V being the final velocity
Xo being the original position
Vo being the original velocity
a being acceleration

A:
X = 40.0m + (5.00m/s)(0.400s) + (0.5)(-9.8m/s^2)(0.400s)^2
X = 41.216 meters

B:
V = 5.00m/s + (-9.8 m/s^2)(0.400s)
V = 1.08m/s

C:
X = 40.0m + (5.00m/s)(1.35s) + (0.5)(-9.8m/s^2)(1.35s)^2
X = 37.81975 meters

D:
V = 5.00m/s + (-9.8m/s^2)(1.35s)
V = -8.23m/s

E:
You need to make X = 0 and solve for time.
0 = 40 + 5(t) + 1/2(-9.8) t^2
Using the Quadratic formula
t = (-b +- sq(b^2 - 4 a c)) / 2a

a = -4.9
b = 5
c = 40

one of the answers is negative so we don't use that number
the other answer is around 9.166 seconds

F:
Using this time we can imput it into the velocity formula.
V = Vo + a t
v = 5 + -9.8 (9.166)
v = -84.83 m/s

G:
To find the greatest height, we need to find the time it takes for the velocity to reach 0 m/s.
So we input 0 into the velocity formula as the final velocity.
0 = 5 + ( -9/8)t
t = 0.510204 seconds

plug this into the position formula
x = 40 + 5(.510204) + 1/2 (-9.8) (.510204)^2
x = 41.2755 meters

Answer 2

A.
s = s0 + v0t + ½at^2
s = 40 + 5*0.4 - 4.9*0.4^2
s = 41.216 m

B.
v = v0 + at
v = 5 - 9.8*0.4
v = 1.08 m/s

C.
s = 40 + 5*1.35 - 4.9*1.35^2
s = 37.82 m

D.
v = 5 - 9.8*1.35
v = -8.23 m/s

E. How many seconds after its release will the bag strike the ground?
s = s0 + v0t + ½at^2
4.9t^2 - 5t - 50 = 0
t = (5 ± √(25 + 980))/9.8
t = (5 ± √(1005))/9.8
t = (5 ± 31.702)/9.8
t = 3.7451 s

F: With what magnitude of velocity does it strike?
v = 5 - 9.8*3.7451
|v| = 31.702 m/s

G: What is the greatest height above the ground that the sandbag reaches?
v^2 - v0^2 = 2a(s - s0)
s = (v^2 - v0^2)/2a + s0
s = (0 - 5^2)/(-2*9.8) + 40.0
s = 41.276 m

Answer 3

good luck