Physics, 30.01.2020 07:45 jellyangie1

Acannonball is fired across a flat field at an angle of 43 degrees with an initial speed 32 m/s and height of 12 m.

1)write a set parametric equations for the motion of the cannon ball
2)determine how long the cannon ball was in the air
3)determine how far the cannon ball traveled in the air
4) determine when the cannon ball reaches its maximum height
5) determine the maximum height reached by the cannon ball

Acannonball is fired across a flat field at an angle of 43 degrees with an initial speed 32 m/s and

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Answer from: kenoknox

1) $x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2$

The initial data of the projectile are:

y_0 = 12 m is the initial height

v_0 = 32 m/s is the initial speed of the projectile, so its components along the x- and y- directions are

$v_{0x} = v_0 cos \theta = (32 m/s)(cos 43^{\circ})=23.4 m/s\\v_{0y} = v_0 sin \theta = (32 m/s)(sin 43^{\circ})=21.8 m/s$

The motion of the cannonball along the x-direction is a uniform motion with constant speed, while on the y-direction it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 downward. So, the two equations of motion of the projectile along the two directions are:

$x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2$

2) 4.94 s

To determine how long the cannon ball was in the air, we need to find the time t at which the cannonball hits the ground, so the time t at which y(t)=0:

0=12+21.8t-4.9 t^2

Solving the equation with the formula, we have:

$t_{1,2}=\frac{-21.8\pm \sqrt{(21.8)^2-4(-4.9)(12)}}{2(-4.9)}$

which has two solutions:

t = -0.50 s

t = 4.94 s

Discarding the first solution which is a negative time so it has no physical meaning, the correct solution is

t = 4.94 s

3) 115.6 m

To determine how far the cannonball travelled, we need to find the value of the horizontal position x(t) when the ball hits the ground, at t=4.94 s. Substituting this value into the equation of motion along x, we find:

$x=v_{0x}t=(23.4 m/s)(4.94 s)=115.6 m$

4) 2.22 s

The cannonball reaches its maximum height when the vertical velocity becaomes zero.

The vertical velocity at time t is given by

$v_y(t)= v_{0y} -gt$

where

g = 9.8 m/s^2 is the acceleration due to gravity

Substutiting v_y(t)=0 and solving for t, we find

$t=\frac{v_{0y}}{g}=\frac{21.8 m/s}{9.8 m/s^2}=2.22 s$

5) 36.2 m

The maximum height reached by the cannon is equal to the vertical postion y(t) when the vertical velocity is zero, so when t=2.22 s. Substituting this value into the equation of the vertical motion, we find:

$y(t)=y_0 + v_{0y}t-\frac{1}{2}gt^2=12+(21.8)(2.22)-(4.9)(2.22)^2=36.2 m$

Answer from: Quest

Answer

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

explanation

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

Answer from: Quest

at right angles to each other

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