Answer:
The options are not shown, so i will answer in a general way.
Suppose the case where the forces act in opposite directions, then we need to subtract the forces, and we know that the magnitude of the resultant force will be:
60N - 50N = 10N
Now, suppose the case where both forces act in the exact same direction, in that case, we will add the forces to get:
60N + 50N = 110N
Then the only range of forces that we can get in this system, are the forces such:
10N ≤ F ≤ 110N
Any resultant force outside that range is not possible in this situation.
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)
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