The speed of the rock at 20 m is 34.3 m/s
Explanation:
We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:
where
:
is the initial potential energy
is the initial kinetic energy
is the final potential energy
is the final kinetic energy
The equation can also be rewritten as follows:
where:
m = 100 kg is the mass of the rock
is the acceleration of gravity
is the initial height
u = 0 is the initial speed (the rock starts at rest)
is the final height of the rock
v is the final speed when h = 20 m
And solving for v, we find:
Learn more about kinetic energy and potential energy here:
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It’s a force which acts on an object without coming physically in contract with it.
Answer:
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
Explanation:
initial veetical speed V₀y=0
Horizontal speed Vx = Vx₀= 3.80m/s
Vertical drop height= 3.90m
Let Vy = vertical speed when it got to the water downward.
g= 9.81m/s² = acceleration due to gravity
From kinematics equation of motion for vertical drop
Vy²= V₀y² +2 gh
Vy²= 0 + ( 2× 9.8 × 3.90)
Vy= √76.518
Vy=8.747457
Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below
V= √Vy² + Vx²
V=√3.80² + 8.747457²
V=9.537m/s
The angle can also be calculated as
θ=tan⁻¹(Vy/Vx)
tan⁻¹( 8.747457/3.80)
=66.52⁰
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
