Answer:
a) t = 2.64s
b) Vf = -28.7m/s
Explanation:
If the balloon is descending, the velocity is -2.3m/s. So the equation to describe the postion of the falling camera is:

Solving for t, we get:
t1 = -3.1s and t2 = 2.64s We discard the negative time and use the positive one.
The velocity of the camera will be:
Vf = Vo - g*t = -2.3 - 10*2.64 = -28.7m/s
Answer:
applied force
Explanation:
any force where you push or pull is always applied force.
Answer:
(97°F − 32) × 5/9 = 36.111°C
Explanation:Hope this helped
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s
Let h₁ = the height at time t, for t ≤ 3.98 s
Let h₂ = the height at time t > 3.98 s
Motion for t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s
Motion for t > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m
The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m
Answer: 931.4 m (nearest tenth)