A motorbike is traveling to the left with a speed of 27.0\,\dfrac{\text m}{\text s}27.0 s m 27, point, 0, start fraction, star
1 answer:
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Answer:
a)Initial speed of the projectile = 196.2 m/s
b)Maximum altitude = 490.5 m
c) Range of projectile = 3398.28 m
d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
Explanation:
Time of flight of a projectile is given by the expression,
Here θ = 30° and t = 20 s
a)
Initial speed of the projectile = 196.2 m/s
b) Maximum altitude is given by
Maximum altitude = 490.5 m
c) Range of projectile is given by
Range of projectile = 3398.28 m
d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s
Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s
We have equation of motion s = ut + 0.5 at²
Horizontal motion
u = 169.91 m/s
a = 0 m/s²
t = 15 s
Substituting
s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m
Vertical motion
u = 98.1 m/s
a = -9.81 m/s²
t = 15 s
Substituting
s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m
Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
Answer:
V₂ = 1.5 m/s
Explanation:
given,
speed of the first piece = 6 m/s
speed of the third piece = 3 m/s
speed of the second fragment = ?
mass ratios = 1 : 4 : 2
fragment break fly off = 120°
α = β = γ = 120°
sin α = sin β = sin γ = 0.866
using lammi's theorem
A,B and C is momentum of the fragments
4 x V₂ = 2 x 3
V₂ = 1.5 m/s
