Question

What is the landing point?

A projectile is fired due east with initial speed 100 ft/s from the top of a hill 20 ft above the ground at an angle `pi/3` above the horizontal. The object has a mass of 1 slug. There is a southerly Magus force of 4 pounds.

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the solution is (24.3,174.2,0). I don't know how to start it?

Answer 1

Unable to get posted solution.

Explanation:

Let us define three dimensional coordinate system with origin located at the ground level below the point of projection. The projectile has three motions. Vertically up `hatz`, Horizontal `hatx` and Southerly `hat y`. As all three directions are orthogonal to each other, each can be treated separately.

Vertical motion.
To calculate the time of flight `t` we use the kinematic expression

`s=s_0+ut+1/2at^2` ........(1)

Taking `g=32\ fts^-2`, noting that gravity acts in the downwards direction, remembering that when the projectile hits ground its height is `z=0`, and inserting given values we get

`0=20+[100sin(pi/3)]t+1/2(-32)t^2`
`=>0=20+[100sqrt3/2]t-16t^2`
`=>8t^2-25sqrt3t-10=0`

Found roots of this quadratic equation using in built graphics tool as

`t=-0.222 and 5.635\ s`.

Ignoring the `-ve` root as time can not be negative we have time of flight

`t=5.635\ s` ........(2)

Horizontal motion.
Distance traveled `x` during time of flight, with initial horizontal velocity `=100cos(pi/3)=50\ fts^-1`

`x=50xx5.635=281.75\ ft`

Southerly motion.
Given mass of projectile `=1\ slug~~32.17\ lb`
Force is given `=4\ lb`
From Newton's Second Law of Motion we get southerly acceleration `a` as

`F=ma`
`=>a=4/32.17\ fts^-2`

Using (1) we get southerly displacement as

`y=-(0xx5.635+1/2xx4/32.17xx(5.635)^2)`
`y=-1/2xx4/32.17xx(5.635)^2`
`y=-1.97\ ft`

I found `(281.75,-1.97, 0)`