Answer to Question #136717 in Classical Mechanics for Ysa Arcena

Question

A sniper holding a Barret M82 Sniper Rifle fires a bullet with a speed of 853 m/s at an
angle of 35° above the horizontal.

a. What height does the bullet reach?
b. How long is the bullet in the air?
c. What is the horizontal range?

Answer 1

As per the given question,

Speed of the bullet $(u)=853 m/sec$

Angle $(\theta)=35^\circ$

a) Height reach by the bullet $(H_{max})=\frac{u^2\sin^2\theta}{2g}$

Now substituting the values, $(H_{max})=\frac{853^2\sin^235^\circ}{2g}$

$\Rightarrow (H_{max})=\frac{727609\times 0.329}{2\times 9.8}m$

$\Rightarrow (H_{max})=12213.43m$

b) Let the bullet stay till T time,

$(t)=\frac{2v\sin\theta}{g}$

Now, substituting the values,

$(t)=\frac{2\times853\times \sin35}{9.8}sec$

$\Rightarrow (t)=99.84sec$

c) Let the horizontal range of the bullet is (R)

$(R)=\frac{v^2\sin2\theta}{g}$

Now, substituting the values,

$(R) =\frac{853^2\times\sin 70}{9.8}m$

$\Rightarrow (R)=69768.24m$