How am I suppose to make an equation without numbers?

Question

A cannon sits near a castle and is aimed away from the castle at an angle above the horizontal.
Find the equation for the time for the cannonball to hit the level ground below in terms of the hheight of the battlement, the megnitude and direction of the initial velocity and the acceleration due to gravity.

can someone explain or show this to me?

Answer 1

-+he is trying to make you mix up equation and realize what is going on .....
when the canon shoots at an angle the cannon ball will travelin an arc til its reaches the level of the canno and then it will vontinue to move till it hit the ground. always keeping it Ux consant.. so the importn factor in this is the veritcal movement. if up is + and down is - then you through an object upwards and you must find out the time it will take it to reach maximum height and the to -H (SAY HEIGH OF CANNON FROM THE GROUND)



so -H = Usin(theta)*t - 1/2gt^2 that leads to the

1/2gt^2 -Usin(theta)*t -H = 0 THAT LEADS TO THE FORM OF

ax^2 - bx-c

wotk out the aljebra and you will find 2 solution
the negative one is not valid therefore time will be eventually known

Answer 2

This is a problem of launch at an oblique angle from horizontal (the angle above the horizontal, that we will call Alpha).
Use these formulaes:
A: angle with horizontal
v0 = initial velocity
s = horiz distance from start point
h = height reached after a given time
t = time (sec)
g = gravity
v = velocity at any point
H = maximum height
L = distance max
We have:
s = v0 * t * cos (A) in m
h = v0 * t * sin(A) - (g * t² / 2)
v = sqr(v0² - (2 * g * h))
L = (v0² / g) * sin (2A)
H = (v0² / 2*g) * sin² (A)
Good luck!

Answer 3

You have to make up your own numbers. Use variables for the angle and the initial velocity. The direction shouldn't affect it, unless you're taking the prevailing wind into account.

Answer 4

I guess this is the std derivation for range.If you wanna learn it IM me
at Jaagrut_live its too laborious to type.Hope you have mic n headphone:)