Projectile Motion Questions _please help!

[squance]

Hey.
My sister would like some help on these physics questions from her Heinemann physics textbook:

1). A softball of mass 250g is thrown with an initial velocity of 16 ms^-1 at an angle theta to the horizontal. When the ball reaches its maximum height, its kinetic energy is 16J.

What is the displacement of the ball after 1.0 s?

2). During training, an aerial skier takes off from a ramp that is inclined at 40.0 degrees to the horizontal and lands in a pool that is 10.0  m below the end of the ramp. If she takes 1.50 s to reach the highest point of her trajectory, calculate:

the speed at which she leaves the ramp.

[vce08]

number 1.

At maximum height the velocity of the ball is only horizontal, there is no vertical velocity.
Thus the horizontal velocity of the ball would be 16= 0.5mv^2, hence V vertical = 11.3m/s

Thus after 1 s the horizontal displacement of the ball would be 11.3 metres, since the horizontal velocity of the ball would not change (no air resistance in VCE physics).

(calculation for total displacement shown below)

number 2

I don't think the 10m pool thing data is relevant.
Using the vertical component of motion at max point, V vertical = 0 and time to reach this is 1.5s
Hence initial vertical velocity can be calculated. v= u + at ,   uvertical = 15m/s

Hence the initial velocity can be calculated using simple trigonometry.
Initial velocity  = 15/sin40 = 23.3m/s

[squance]

Thanks
My sister appreciates the response.
But the answers in the book are

1). 13 m at an angle of 30 degrees
2). 22.9 m s^-1

Not sure if there are errors in the book but my sister looked at your working and it seems logic to her.

Thanks again.

[vce08]

number 1 is cause they found out the total displacement not just the horizontal displacement.
You would then have to proceed to working with the vertical component of motion.
From part 1 you would know that the horizontal velocity is 11.3m/s, hence using pythagoras you can find out the initial vertical velocity to be 11.3m/s. After 1 second you can find out the vertical displacement using the formula x = ut + 0.5at^2, hence x = 11.3 - 4.9 = 6.4m
Thus using pythagoras once more using the vertical and horizontal displacements you can find the total displacement which is
root(6.4^2 + 11.3^2) = 13m.
to find the angle you would use arctan(6.4/11.3) = 30 degrees.

number 2 is cause they used g = 9.8m/s^2.

[danieltennis]

Question 2

Consider the vertical component of velocity

where V = the speed she leaves the ramp

[squance]

Thanks Daniel!
It makes sense now

[squance]

Thanks again Table.

[squance]

Oh! And this is a question from the 2009 Checkpoints.

A speeding motorbike travels past a stationary car at a speed of 35 m s^-1 . The police car starts accelerating immediately at 4 m s^-2, and keeps accelerating at this rate until it has passed the bike.

How far does the police car travel before it overtakes the motorbike?

The answer is apparently 613 m but my sister can't seem to get it.

[/0]

The distance the bike travels is

The distance the policeman travels is





.

So when the police car catches up it has traveled

[squance]

Thanks /0!

[Mao]

I've seen a few my-sister questions..
Why does she not register an account herself? I'm sure she'll be welcomed.

[squance]

I've seen a few my-sister questions..
Why does she not register an account herself? I'm sure she'll be welcomed.

My sis doesn't use the internet a lot and she told me she can't be stuffed registering an account.
LOL
And if she happens to use the internet, the first page that comes up is VN...with my account already logged in

[kurrymuncher]

I wish I had a sister like you