Explanation:
For projectile motion, use constant acceleration equation:
Δx = v₀ t + ½ at²
where Δx is the displacement,
v₀ is the initial velocity,
a is the acceleration,
and t is time.
Both objects are projected upward with velocity u. The second object is thrown after a time t₀.
For the first object:
Δx = u t + ½ (-g) t²
Δx = u t − ½g t²
For the second object:
Δx = u (t−t₀) + ½ (-g) (t−t₀)²
Δx = u (t−t₀) − ½g (t−t₀)²
Assuming the objects meet, the displacements will be equal:
u t − ½g t² = u (t−t₀) − ½g (t−t₀)²
u t − ½g t² = u (t−t₀) − ½g (t² − 2tt₀ + t₀²)
u t − ½g t² = u t − u t₀ − ½g t² + g tt₀ − ½g t₀²
0 = -u t₀ + g tt₀ − ½g t₀²
0 = -u + g t − ½g t₀
g t = u + ½g t₀
t = u/g + t₀/2
Yes! Although the object has potential energy, the object is not moving and therefore has no speed. :)
The answer is <span>The power will decrease to 75 percent of its previous value
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Answer:
v= 4.9 m/s in east direction ( or v=4.9 m/s * i )
Explanation:
Since the passenger moves along the train ( in the north direction relative to the train) then if the train is traveling east relative to the ground , the passenger will also travel to the east relative to the ground .
Then the magnitude of the velocity will be
v= 1.40 m/S + 3.5 m/s = 4.9 m/s
v= 4.9 m/s
and the direction will be east ( if the x axis represents east direction and y-axis the north direction , then the velocity vector will be v=4.9 m/s * i )
