Answer:
The take-off speed is 41.48 
Explanation:
Given :
Range
m
Projectile angle
13°
From the formula of range,

Find the velocity from above equation,

( ∵
)

Therefore, the take-off speed is 41.48
Answer:
a) F = 527.65 N, Force applied is upwards.
b)F = - 527.65 N, where, negative sign depicts Force is applied downwards.
Explanation:
Data provided:
Weight of the firefighter = 756 N
Mass of the firefighter = 756/9.8 = 77.14 Kg
Acceleration, a = 2.96 m/s²
a) In the absence of the pole the firefighter would have been moving down with an acceleration of 9.8 m/s (i.e the acceleration due to the gravity), but due to the presence of the pole the acceleration of the firefighter has been reduced. thus, a force is applied by the pole on the firefighter to reduce the acceleration.
therefore, we have
F = ma(net) = 77.14 × (9.8-2.96) = 527.65 N, Force applied is upwards.
B) According to the Newton's third law, the force will be equal and opposite to the force in the part a)
thus, we have
F = - 527.65 N
The net force must be zero
This is in accordance to Newton's first law, which states that any object in motion will remain in motion and any object at rest will remain at rest unless acted upon by an unbalanced force. An unbalanced force is one where the net force is not zero. If no unbalanced force is applied to a moving object, it will keep moving forever. The reason that we do not observe this in our daily lives is due to friction acting as the unbalanced force.