help im kinda confused on this

[worldly plank]

i have no idea how am i supposed to start

[slate glen]

Isn’t this physics?

Can we solve this with just algebra or has to be calculus

[inland lance]

ok.. i might help you

remember the projectile equation?

[worldly plank]

i think its okay cuz i kinda find a path

so um the H which is height

[slate glen]

I know how to solve with physics but not through some fancy math

[worldly plank]

is 64 meters

i literally asked my teacher

he kinda suffered too lol

[inland lance]

what is 64?

[worldly plank]

the height

ill say the length of y axis

[inland lance]

u mean max height?

[worldly plank]

yes

its okay tho ive found my answer

im going to have an exam tomorrow

so yeah

[inland lance]

ok..

i think there is a formulat i learned

i.e

range $R=u\sqrt{\frac{2h}{g}}$

[gloomy stump]

pratham

[worldly plank]

oh i dont use that

so what we are finding is their initial velocity

[inland lance]

y?

yes

[worldly plank]

the range is can be found by looking at the diagramn

[inland lance]

yes ofc

[worldly plank]

aight i gotta went busy with other chapters

cya later if we can

[inland lance]

but we can find other quantities by substituting R

r=100 ryt?

ok.. i might find max height by using projectile eq

[worldly plank]

go search Curvilinear Motion

all formulas are there

[inland lance]

$y=\tan{\theta}x-\frac{g}{2u^2 \cos^2 \theta}x^2$

[gloomy stump]

pratham

[inland lance]

but i need to translate 4 units up

do you get it?

height of ramp is 4

just add 4 to eq

[worldly plank]

yup

[inland lance]

you may differentiate to get max height

[worldly plank]

well tho thanks for the advice but i had to continue other chapter

[inland lance]

then substitute here

alright will you try now?

[inland lance]

k np

[worldly plank]

alright thanks

[inland lance]

do whenever you have time