((THIS IS WORTH 50 POINTS))
2 answers:
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Answer:
just before landing the ground
Explanation:
Let the velocity of projection is u and the angle of projection is 30°.
Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.
initial horizontal component of velocity, ux = u Cos 30
initial vertical component of velocity, uy = u Sin 30
Time of flight is given by
Final horizontal component of velocity, vx = ux = u Cos 30
Let vy is teh final vertical component of velocity.
Use first equation of motion
vy = uy - gT
vy = - u Sin 30
The magnitude of final velocity is given by
v = u
Thus, the velocity is same as it just reaches the ground.
Answer:
v₁ = -0.8087 m / s
Explanation:
To solve this problem we can use the conservation of momentum, for this we define a system formed by the man, the skateboard and the brick, therefore the force during the separation is internal and the momentum is conserved
Initial instant. When they are united
p₀ = 0
Final moment. After throwing the brick
= (m_man + m_skate) v1 + m_brick v2
the moment is preserved
p₀ = p_{f}
0 = (m_man + m_skate) v₁ + m_brick v₂
v₁ = -
the negative sign indicates that the two speeds are in the opposite direction
let's calculate
v₁ = -
v₁ = -0.8087 m / s