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1
Answer:
hope it helps uh :)
Explanation:
The Horizontal distance covered by the projectile when it is projected at an angle 45 degrees. Because, Range = u²sin2θ/g. sin2θ has a maximum value when theta is 45 degrees.
Firing of a bullet is a projectile.
Given, Horizontal range of the bullet = 10km.
From the relation,
\begin{gathered}\large {R = \frac {u^2 sin2 \theta} {g}} \\ \\\end{gathered}
R=
g
u
2
sin2θ
We have, sin2θ = 1, g = 9.8 m/s²
\begin{gathered}R = 10 \times {10}^{3} = \frac{ {u}^{2} \times 1}{g} \\ \\ \implies \: 10000 \times 9.8 = {u}^{2} \\ \\ \implies \: u = \sqrt{98000} \\ \\ \implies \: u \: = 313 \: m {s}^{ - 1}\end{gathered}
R=10×10
3
=
g
u
2
×1
⟹10000×9.8=u
2
⟹u=
98000
⟹u=313ms
−1
Maximum Height :
The maximum vertical distance reached a projectile when it's velocity becomes zero.
\begin{gathered}H_{max} = \frac{u ^{2} sin ^{2} \theta} {2g} \: \\ \\\end{gathered}
H
max
=
2g
u
2
sin
2
θ
\begin{gathered}H_{max} = \frac{98000 \times (\frac{1}{ \sqrt{2}}) ^{2} }{2 \times 9.8} \\ \\ H_{max} = \frac{10000}{4} = 2500m ≈ 2492 m\end{gathered}
H
max
=
2×9.8
98000×(
2
1
)
2
H
max
=
4
10000
=2500m≈2492m
The velocity of projection and the maximum height attained by the bullet are 313 m/s, 2492 m