Question

A stone is projected from a point O on horizontal ground with speed V m s−1 at an angle θ above the
horizontal, where sinθ = 3
5
. The stone is at its highest point when it has travelled a horizontal distance
of 19.2 m.
(i) Find the value of V. [3]
After passing through its highest point the stone strikes a vertical wall at a point 4 m above the ground.
(ii) Find the horizontal distance between O and the wall. [4]

Answer 1

i solved i) but i am stuck on ii)

Answer 2

U know
Horizontal component of initial velocity =v*cos theta

Answer 3

Vertical component of initial velocity=v*sin theta

Answer 4

yes

Answer 5

Use for y component
y=v*sin theta*t-0.5gt^2
Given y=4m

Answer 6

shouldnt we take into accout the maximum height

Answer 7

Ya

Answer 8

x=v*cos theta*t=?

Answer 9

I think u know initial velocity frm i)

Answer 10

for i) at max height v=0 m/s Vertical velocity vsinθ-gt=0 and horizontal displacement of vcosθt=19.2 t=19.2/vcosθ plug that back into the vertical velocity vector. vsinθ-g(19.2)/vcosθ=0 cosθ=4/5 and sinθ=3/5. solve for v, v=20m/s^2.

Answer 11

sorry m/s

Answer 12

Ok

Answer 13

20cosθt=19.2
t=19.2/16=1.2 seconds
20sinθt-5t^2=hmax
hmax=7.2m

Answer 14

ii) u=0m/s a=gm/s^2 s=4m v=vm/s use v^2=u^+2as v=4root5 m/s

Answer 15

vertical displacement:7.2-vsinθt-1/2gt^2=4
(7.2-4)+4root5*3/5*t-5t^2=0

Answer 16

t=1.499 seconds

Answer 17

Ya

Answer 18

Now x=v*cos theta*t=?

Answer 19

when i checked out my answers the distance from O to the wall is incorrect but 4root 5 the speed is correct.

Answer 20

I hv little confusion!!!

Answer 21

U wrote
Vertical displacement 7.2. Why?!!!

Answer 22

Can u use
y=v*sin theta*t-0.5gt^2
Given y=4m
v=20
t=?

Answer 23

7.2 m is the maximum height

Answer 24

I c

Answer 25

why is speed equal to 20m/s shouldnt it equal to 4root5

Answer 26

That wrong to use
h=7.2-ut-0.5gt^2

Answer 27

okay then have we chosen speed to be equal to 20m/s

Answer 28

U calculated the initial velocity of projection v=20m/s
Right?

Answer 29

yes thats the initial as it projects upwards till it reaches max height of which then velocity equals zero then velocity increases again

Answer 30

yes i did.

Answer 31

But this is initial velocity of projection v=20m/s

Answer 32

Now can u tell me t=?
Frm my equation

Answer 33

2 seconds

Answer 34

Now use
x=v*cos theta*t=?

Answer 35

v=32m/s

Answer 36

v=32m/s?!!!

Answer 37

Did u get it???

Answer 38

i still don't get why the intital velocity of projection should be the same after the particle reaches maximum height.

Answer 39

At the instant when the stone hits the wall the horizontal component of the stone’s velocity is halved in
magnitude and reversed in direction. The vertical component of the stone’s velocity does not change
as a result of the stone hitting the wall.
(iii) Find the distance from the wall of the point where the stone reaches the ground.

Answer 40

Actually velocity of projection is not equal to the velocity at highest point!!

Answer 41

U know any stone going upward and its velocity decreases

Answer 42

But my question is... why u r using the velocity of highest point?

Answer 43

Actually ur initial velocity of projection is v

Answer 44

X comonent of v is v*cos theta. And this component of velicity is responsible for displacement along x axis

Answer 45

Y component of v is v*sin theta. And this component is responsible for displacement along y axis

Answer 46

Our actual equation is
h=ut-0.5gt^2
But we write this equation here as
y=v*sin theta*t-0.5*g*t^2

Answer 47

I use
u=v*sin theta
Because of displacement along y axis

Answer 48

I wanna clear ur confusion!!!

Answer 49

its crystal clear now thanks @shamim.

Answer 50

U r most welcome!!!!!