Large amounts of water that quickly infiltrate soil on a sloped surface can

A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra

seraphim [82]

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:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial speed of the projectile

$\theta$ is the angle of projection

$g=9.8 m/s^2$ is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

$\theta=60^{\circ}$ (angle)

Solving the equation, we find the horizontal range:

$d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m$

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:

$v_y^2 - u_y^2 = 2as$

where

$v_y$ is the vertical velocity of the bullet after having covered a vertical displacement of $s$

$u_y$ is the initial vertical velocity

$a =-g=$ is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

$u_y = u sin\theta$

Also, the maximum height s is reached when the vertical velocity becomes zero,

$v_y =0$

Substituting into the equation and re-arranging for s, we find the maximum height:

$s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

3 years ago

HELP PLEASE 100 POINTS!

erica [24]

Yes the relationship between them is direct linear ..

4 0

3 years ago

Read 2 more answers

You grab a hot cookie sheet out of the oven this is an example of ______ energy transfer.

Maru [420]

I would say Conduction because you are touching the cookie sheet, even though it is hot (so heat) you are physically touching it so it would not be radiation

6 0

3 years ago

1. Find the charge if the number of electrons is 4 x 10-18

Marianna [84]

22. The answer is 22.

4 0

3 years ago

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0

3 years ago