1) The horizontal range of the bullet is 884 m
2) The maximum height attained by the bullet is 383 m
Explanation:
1)
The motion of the bullet is a projectile motion, which consists of two separate motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:
[tex]d=\frac{u^2 sin 2\theta}{g}[/tex]
where
u is the initial speed of the projectile
[tex]\theta[/tex] is the angle of projection
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
For the bullet in the problem, we have
u = 100 m/s (initial speed)
[tex]\theta=60^{\circ}[/tex] (angle)
Solving the equation, we find the horizontal range:
[tex]d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m[/tex]
2)
To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:
[tex]v_y^2 - u_y^2 = 2as[/tex]
where
[tex]v_y[/tex] is the vertical velocity of the bullet after having covered a vertical displacement of [tex]s[/tex]
[tex]u_y[/tex] is the initial vertical velocity
[tex]a =-g=[/tex] is the acceleration (negative, since it points downward)
The vertical component of the initial velocity is given by
[tex]u_y = u sin\theta[/tex]
Also, the maximum height s is reached when the vertical velocity becomes zero,
[tex]v_y =0[/tex]
Substituting into the equation and re-arranging for s, we find the maximum height:
[tex]s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
#LearnwithBrainly
