The equation of trajectory of a projectile is y=10x-(5/9)x^2 if we assume g=10m/s^2, the range of projectile in meters is?
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Answer:
v_squid = - 2,286 m / s
Explanation:
This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.
Initial moment. Before expelling the water
p₀ = 0
the squid is at rest
Final moment. After expelling the water
= M V_squid + m v_water
p₀ = p_{f}
0 = M V_squid + m v_water
c_squid = -m v_water / M
The mass of the squid without water is
M = 9 -2 = 7 kg
let's calculate
v_squid = 2 8/7
v_squid = - 2,286 m / s
The negative sign indicates that the squid is moving in the opposite direction of the water