A projectile is fired from the origin O at an angle
of 45° from the horizontal . At the highest
point p of its trajectory the radial and
transverse components of its accleration in terms
of the gravitational accleration g are ?
Answers
Answer:
A projectile is fired from a gun at an angle of 45
o
with the horizontal and with a speed of 20 m/s relative to ground. At the highest point in its flight the projectile explodes into two fragments of equal mass. One fragment comes at rest just after explosion. How far from the gun does the other fragment land, assuming a horizontal ground? Take g=10m/s
2
?
HARD
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ANSWER
As the first part comes to the rest means, then it will fall under gravity directly.
First of all maximum height,
H=
g
u²sin²θ
H=
2×10
400×(
2
1
)
H=10m.
As initially the full projectile was supposed to land at A but due to explosion first part land at A', so to make the centre of mass lies at A the second part should lie at B.
f they don't break (explode) range OA
OA=
g
u²sin2θ
OA=400×sin45×
10
2
OA=400×
10
sin90°
OA=40m.
Therefore ,
AA
′
=
2
OA
AA
′
=20m.
As the first part is at 20 m from A in left side being equal mass the second part will be at 20 m right to A.
From gun it is OB=40+20m=60m.
so gun does the other fragment land at 60m
Explanation:
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