Progectiles

[zenith trench]

Yk for this q
Why woundt you solve vertically with s=vt

[mighty bear]

because

gravity is acting downards

in the vertical plane

so you can't do that since theres a net acceleration

s=vt only applies for the horizontal motion given that no external forces are acting in that direction bu theres friction

if the horizontal velocity was constant then sure u can use s = vt

so

.

for part a)
if something is projected horizontally, its vertical velocity is 0

the time of 1 second
is the time i took for it travel 2m to the right then 1.2m down

time time under gravity
s = 1.2
a = g
u = 0 (since we are writing vertical suvat)

s = ut+1/2at^2
so you solve for t

1 - what you got for t is the time for whic h it was in contact iwth the surface

for part b)

what i would do is

you work out the change in kinetic energy

we all know that energy is conserved so if you see how much energy was lost

that means that energy was lost due to the resitive forces, i.e. friciton

and work done = force * distance
so
energy lost = friction force * 2

then plug it in
F = u * R

OR
you have the time for which it was in contact with
so write suvat, taking the right as positive
to work out the new deceleration
(since only resitive forces are acting)
and plug into F = ma to get resistive force