Question

a particle projected at 40 degree to the horizontal reaches its greatest height after 3 seconds. calculate the speed of its projection

Answer 1

Do you have any idea how to solve this problem ?

Answer 2

Horizontal component-vcosѲ
Vertical component-vsinѲ
At maximum height v=0m/s a=10m/s^2 t=3 seconds
v=u-at, 0=u-(10*3), u=30 m/s

Answer 3

horizontal component is constant as no force by gravity is acting on it
30*cos(40)=22.98 m/s

Answer 4

0=VsinѲ-at
V*sin40=10*3
0.643V=30
V=46.6

Answer 5

I think you can get the answer by considering the vertical motion only

Answer 6

ah sorry, i just saw your last message - yes you got it !

Answer 7

thanks for your time :)

Answer 8

welcome

Answer 9

A canon fires a shot at 38 degree above the horizontal. the initial speed of the cannonball is 70m/s. calculate the distance between the canon and the point where the cannonball lands. given that the two points are at the same horizontal level.

Answer 10

Is it supposed to be the horizontal distance

Answer 11

You could work out the time similarly to the last problem, but setting final vertical speed equal and opposite to initial vertical speed
That would let you work out the time of flight

Answer 12

Then it would be a simple matter to use the time and the horizontal speed to work out the distance covered along the horizontal

Answer 13

Vx=v*cosѲ=70*cos 38
vy=v*sinѲ=70*sin 38
i think we are going to write down two equation which are in terms of both V(final speed) and time of the whole journey

Answer 14

I have used the fact that when the cannonball hits the ground again, it's vertical speed will just be minus its initial vertical speed

Answer 15

and used v=u-gt

Answer 16

no time is provided

Answer 17

you can use v= u-gt to work out the time

Answer 18

we know u is Vinitialxsin38 and u is -Vinitalxsin38

Answer 19

that x means 'multiply', not x axis

Answer 20

argh ! another typo !
we know u is Vinitial x sin38 and v is -Vinitial x sin38

Answer 21

we know this from vsquared=usquared + 2as because s here (net vertical displacement) is zero
it's basically conservation of energy applied to the vertical motion

Answer 22

okay u(vertical)=vsin(Ѳ)
displacement vertical=vsinѲ-1/2gt^2=70*sin38-1/2(10)t^2
Horizontal Displacement=vcosѲ*t
but isnt the equation v^2=u^2+2as involved with objects travelling in a straight line.

Answer 23

It's okay to apply it to the vertical motion independently - but we only needed it to tell us that the magnitued of vertical speed is the same at the end as it was at the beginning

Answer 24

remember h=0 at our start and end points

Answer 25

okay i agree with you that displacement is zero, okay but how is the magnitude of vertical speed the same isn't gravity acting on it downwards.

Answer 26

if you throw a ball vertically up with speed v, it starts upwards, decelerates to a stop, turns round, picks up speed again and gets back to you with the same speed v but in the opposite direction
the same is true for the vertical motion of any projectile (ignoring air resistance)

Answer 27

so if your cannonball heads up with inital vertical speed vsintheta, it will hit the ground with speed -vsintheta

Answer 28

vertical speed, that is

Answer 29

cannon ball takes the same time and speed to travel upwards and vice versa when moving backwards to the ground

Answer 30

yes that's right

Answer 31

but v has a negative sign when moving backwards

Answer 32

yes

Answer 33

can i ask another question, i am sorry i am asking too much

Answer 34

it's alright, i have time at the mometn

Answer 35

thanks man, how should the diagram of the projectile be.

Answer 36

well those projectiles always move on a curve that is a parabola

Answer 37

something like this for the cannonball problem