M2 questions

A particle is projected with a speed of 21ms^-1 at an angle of elevation of 60 degrees. Find its speed and direction of motion after 1 second, 2 seconds and 3 seconds.

A ball is thrown from O with a speed of 28ms^-1 at an angle of elevation of 60 degrees. it hits a wall which is 5m horizontally from O. By modelling the ball as a particle calculate the height above O at which the ball hits the wall.

I have only just started M2 so am likely to get stuck. can anyone help??

Reply 1

Christophicus

18

The Chameleon

A particle is projected with a speed of 21ms^-1 at an angle of elevation of 60 degrees. Find its speed and direction of motion after 1 second, 2 seconds and 3 seconds.

A ball is thrown from O with a speed of 28ms^-1 at an angle of elevation of 60 degrees. it hits a wall which is 5m horizontally from O. By modelling the ball as a particle calculate the height above O at which the ball hits the wall.

I have only just started M2 so am likely to get stuck. can anyone help??

please help.

Reply 2

El Stevo

11

put the initial velocity into vector form.

Reply 3

The Chameleon

OP

I still don't understand.

Reply 4

Gaz031

15

The Chameleon

A particle is projected with a speed of 21ms^-1 at an angle of elevation of 60 degrees. Find its speed and direction of motion after 1 second, 2 seconds and 3 seconds.

A ball is thrown from O with a speed of 28ms^-1 at an angle of elevation of 60 degrees. it hits a wall which is 5m horizontally from O. By modelling the ball as a particle calculate the height above O at which the ball hits the wall.

I have only just started M2 so am likely to get stuck. can anyone help??

Resolve the initial velocity into vertical and horizontal components. Then consider horizontal and vertical movement seperately.
When you want to find overall speed you can use simple pythagoras to find the magnitude. The trigonomic ratios can be used to find the angle.

Reply 5

The Chameleon

OP

Gaz031

Resolve the initial velocity into vertical and horizontal components. Then consider horizontal and vertical movement seperately.
When you want to find overall speed you can use simple pythagoras to find the magnitude. The trigonomic ratios can be used to find the angle.

do you do pythagoras on the speeds or the distances?

Reply 6

Gaz031

15

The Chameleon

do you do pythagoras on the speeds or the distances?

Using a^2 + b^2 = c^2 gives you the magnitude.
Do you want to find the magnitude of the speed or of the distance?

Reply 7

The Chameleon

OP

Here's the question.

A particle is projected with a speed of 21ms^-1 at an angle of elevation of 60 degrees. Find its speed and direction of motion after 1 second, 2 seconds and 3 seconds.

Reply 8

The Chameleon

OP

I'm really sorry, but I don't have the M2 book and I wouldn't be pestering you if I had.
All I get is photocopied sheets of the exercises which doesn't help if you have no book and a bad teacher.

Reply 9

Gaz031

15

The Chameleon

Here's the question.

A particle is projected with a speed of 21ms^-1 at an angle of elevation of 60 degrees. Find its speed and direction of motion after 1 second, 2 seconds and 3 seconds.

You'd want to find the magnitude of the speed at t=1, t=2 and t=3.
The horizontal velocity is a constant. The vertical velocity changes with time.
Find the vertical velocity at time t=1,2,3 then find the magnitude at each time. Draw a triangle with the different components to find the angle.
Will it help if i work through and post a solution or would you prefer to do it?

Reply 10

Newton

a) By applying Newton's laws of motion:

Horizontally there is no accelation. While vertically the particle will experience a downward acceleration. Therefore:

Horizontally: v=u+at AND since a=0 v=u

As this function of velocity of the particle is independent of time, the horizontal component of the velocity at any time, t, will be:

v=21cos60=10.5m(s^(-1))

While the vertical component by applying v=u-gt implies:

v=usin60-gt

Hence:

H=10.5m(s^(-1))
V=21sin60-gt

Therefore at:

t=1: H=10.5 V=8.39

Apply Pythagoras' Theorem to find the magnitude of the velocity i. e. the speed:

S=root((8.39^2)+(10.5^2))=13.4m(s^(-1))

Therefore the speed=13.4(m^s(-1))

To find the direction find the angle between the two components:

tan@=(8.39/10.5) => @=arctan(8.39/10.5)

Therefore:

S=13.4m(s^(-1)) AND @=38.6deg

And so forth.

b) Let sx and sy be the horizontal and the vertical displacements respectively.

sx=5m

For the horizontal component there is no acceleration. Therefore:

sx=vt

Therefore:

t=sx/v

As we are now considering the horizontal displacement, we use the horizontal component of the velocity. Hence:

t=5/28cos60=0.36s

This is the time that it took the ball to reach the wall, and took it to travel 5m horizontally. We now need to consider the height achieved by the particle in this time:

Use s=ut+(1/2)(a)(t^2)

Because the only force acting on it is the gravitational acceleration, and it is acting downwards i. e. against the motion of the particle. So treat acceleration as a negative:

sy=ut-(1/2)(g)(t^2)

Since we are considering the vertical displacement we take the vertical component of the velocity. Therefore:

s=28sin60(0.36)-4.9(0.36^2)

Therefore:

sy=8.09m

Therefore the height=8.09m

MY 150TH POST!!!!!!

Newton.

Reply 11

username9816

14

The Chameleon

1.) A particle is projected with a speed of 21ms^-1 at an angle of elevation of 60 degrees. Find its speed and direction of motion after 1 second, 2 seconds and 3 seconds.

2.) A ball is thrown from O with a speed of 28ms^-1 at an angle of elevation of 60 degrees. it hits a wall which is 5m horizontally from O. By modelling the ball as a particle calculate the height above O at which the ball hits the wall.

In modelling projectile motion, consider a projectile fired at initial speed V, at an angle x to the horizontal, from a defined origin.

Resolving Vertically: u = Vsinx, a = -9.8, t = t.
Let y = Vertical Displacement From Origin.

s = ut + 1/2at^2
---> y = Vsinxt + 1/2(-9.8)(t^2)
---> y = Vsinxt - 4.9t^2

Resolving Horizontally: u = Vcosx, a = 0, t = t
Let z = Horizontal Displacement From Origin

s = ut + 1/2at^2
---> z = Vcosxt

1.) At t = 1:

y = 21(sin60)*1 - 4.9(1)^2 = 13.3 m
z = 21cos60 * 1 = 10.5 m

Let @ = Angle Of Direction Of Motion From The Vertical.
tan@ = opp/adj = 10.5/13.3
---> @ = 38.3 Degrees (3.S.F)

Vertical Motion: u = 21sin60, a = -9.8, t = 1.
v = u + at
---> v = 21sin60 - 9.8(1)
---> v = 8.39 ms^-1
Horizontal Motion: v = Constant 21cos60 = 10.5 ms^-1
Speed = Sq.root (8.39^2 + 10.5^2) = 13.4 ms^-1 (3.S.F)

At t = 2:

y = 21(sin60)*2 - 4.9(2)^2 = 16.8 m
z = 21cos60 * 2 = 21 m

Let @ = Angle Of Direction Of Motion From The Vertical.
tan@ = opp/adj = 21/16.8
---> @ = 51.3 Degrees (3.S.F)

Vertical Motion: u = 21sin60, a = -9.8, t = 2.
v = u + at
---> v = 21sin60 - 9.8(2)
---> v = -1.41 ms^-1
Horizontal Motion: v = Constant 21cos60 = 10.5 ms^-1
Speed = Sq.root ((-1.41)^2 + 10.5^2)) = 10.6 ms^-1 (3.S.F)

At t = 3:

y = 21(sin60)*3 - 4.9(3)^2 = 10.46 m
z = 21cos60 * 3 = 31.5 m

Let @ = Angle Of Direction Of Motion From The Vertical.
tan@ = opp/adj = 31.5/10.46
---> @ = 71.6 Degrees (3.S.F)

Vertical Motion: u = 21sin60, a = -9.8, t = 3.
v = u + at
---> v = 21sin60 - 9.8(3)
---> v = -11.2 ms^-1
Horizontal Motion: v = Constant 21cos60 = 10.5 ms^-1
Speed = Sq.root ((-11.2)^2 + 10.5^2)) = 15.4 ms^-1 (3.S.F)

2.) z = Vcosxt
---> 5 = 28cos60t
---> 5 = 14t
---> t (Hit Wall) = 0.357 s

At t = 0.357 s:
y = Vsinxt - 4.9t^2
---> y = 28sin60(0.357) - 4.9(0.357)^2
---> y = 8.03 m

---> Height Above O (Hit Wall) = 8.03 m (3.S.F)

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