Projectile motion? (1 Viewer)

[P!xel]

There's this question that I dunno how to tackle :|

A person throws a cricket ball so that it's path is given by , where and are measured in metres and in seconds. Find the:

a) Initial velocity and the initial angle of projection.
b) Find the greatest height reached.
c) Find the distance which the ball covered.

This was from MathsInFocus Extension 1 textbook, and it gives answers, but not working.
Could someone show me the working please?
(After I know how to work out part a), I can do the rest )

 

[Timothy.Siu]

probably change x=20t to t=x/20 and sub it into the other equation and u can get them all

 

[k02033]

x(t)=20t, initial horizontal velocity Ux=dx/dt so Ux=20ms^-1
same trick for inital vertical velocity
magnititude of initial velocity U^2=Ux^2+Uy^2 (draw out the 2 component velocities and use a^2=b^2+c^2)
tan(theta)=Uy/Ux (draw out the 2 component velocities and using trig ratio)

 

[Drongoski]

There's this question that I dunno how to tackle :|

A person throws a cricket ball so that it's path is given by , where and are measured in metres and in seconds. Find the:

a) Initial velocity and the initial angle of projection.
b) Find the greatest height reached.
c) Find the distance which the ball covered.

This was from MathsInFocus Extension 1 textbook, and it gives answers, but not working.
Could someone show me the working please?
(After I know how to work out part a), I can do the rest )

Horizontal comp of velocity = dx/dt = 20 m/s
Vert comp of velocity is dy/dt = 48 - 32t m/s = 48 m/s @ t = 0.
Therefore the initial horiz & Vert components of velocity are:
4 x 5 & 4 x 12

By Pythag, initial velocity = 4 x sqrt(5^2 + 12^2) = 4 x 13 m/s
(Note: 5, 12, 13 are a Pythagorean triple)

The initial angle theta is given by : tan(theta) = 48/20 = 12/5

y = 0 when t = 0 and t = 3 ( y(t) is a quadratic in 't' )
Therefore greatest height occurs @ mid-way between 0 & 3 = 1.5
Therefore greatest height = y(1.5) = 16x1.5(3 - 1.5) 16 x 2.25 = 36 m

Horizontal distance covered = 20 m/s x 3 s = 60 m


Is that right ???

 

[P!xel]

I understand the above working out ..

But where abouts did the 4 come from?
Because every line there seems to be a 4x

Edit: I guess it's just a simplification, thanks for the answer.

 

[Drongoski]

I understand the above working out ..

But where abouts did the 4 come from?
Because every line there seems to be a 4x

Edit: I guess it's just a simplification, thanks for the answer.

Oh I'm sorry. The horizontal component of the initial velocity (V say) is 20 m/s and the vertical comp is 48 m/s. So V^2 = 20^2 + 48^2 (rather large numbers) and since it is well known that (5, 12, 13) is a Pythagorean triple, i.e. 5^2 + 12^2 = 13^2; therefore any multiple of of this Pythag triple is also a Pythag triple, e.g. (10, 24, 26), (20, 48, 52), (2000, 4800, 5200) etc. In the case of this problem, since 20 = 4x5, 48 = 4x12, I know V the hypotenuse = 4 x 13. (I like to do simple arithmetic in my head). Hope it is clearer now.

 

[Drongoski]

I'm trying out LaTeX!