M2 projectile- is this even possibru?

I cannot do this whatsoever. I have had numerous attempts but have not been able to solve it.
Q) A particle is projected from a point on level ground with speed u ms^-1 and angle of elevation alpha (a). The maximum height reached by the particle is 42 m above the ground and the particle hits the ground 196 m from its point of projection.
Find the value of alpha and the value of u.

One time 84=0 fell out of the working and found that T=0.5t is imaginary lol.
Any tips? Thanks :tongue:

Reply 1

MM04926412

5

Original post by MathMeister

I cannot do this whatsoever. I have had numerous attempts but have not been able to solve it.
Q) A particle is projected from a point on level ground with speed u ms^-1 and angle of elevation alpha (a). The maximum height reached by the particle is 42 m above the ground and the particle hits the ground 196 m from its point of projection.
Find the value of alpha and the value of u.

One time 84=0 fell out of the working and found that T=0.5t is imaginary lol.
Any tips? Thanks :tongue:

could you not use suvat with s=42 a=-9.8 v=0 to find the starting vertical component, calculate time spent in the air, then use s=d/t to find the horizontal component and then resolve?

Original post by MathMeister

I cannot do this whatsoever. I have had numerous attempts but have not been able to solve it.
Q) A particle is projected from a point on level ground with speed u ms^-1 and angle of elevation alpha (a). The maximum height reached by the particle is 42 m above the ground and the particle hits the ground 196 m from its point of projection.
Find the value of alpha and the value of u.

One time 84=0 fell out of the working and found that T=0.5t is imaginary lol.
Any tips? Thanks :tongue:

I solved it without difficulty.

Post your working.

Original post by MM04926412

could you not use suvat with s=42 a=-9.8 v=0 to find the starting vertical component, calculate time spent in the air, then use s=d/t to find the horizontal component and then resolve?

I really wouldn't. Using the standard equations for maximum height and range yields the tangent of the angle immediately.

Reply 4

MathMeister

OP

10

Original post by Mr M

I solved it without difficulty.

Post your working.

How long?

Reply 5

MM04926412

5

Original post by Mr M

I really wouldn't. Using the standard equations for maximum height and range yields the tangent of the angle immediately.

I didn't do M2, I'm offering the brute force method

Original post by MathMeister

How long?

Pardon?

If you want me to start you off:

Equation for greatest height:

\displaystyle \frac{u^2 \sin^2 \theta}{2g}=42

Equation for range:

\displaystyle \frac{u^2 \sin 2\theta}{g} = 196

So

u^2 \sin^2 \theta = 84g

and

u^2 \sin \theta \cos \theta = 98g

Divide the last two equations.

Reply 7

MathMeister

OP

10

Original post by Mr M

Pardon?...

Show off...
Here is my working out :redface:

... It sound like it would be a good idea if I learnt the formulas for range , greatest height , eqn of path ext...

(edited 9 years ago)

Original post by MathMeister

...

My head doesn't work sideways.

I wouldn't bother learning those equations. They can be derived easily whenever you need them.

Original post by MathMeister

...

I bet you know the equation of the trajectory.

Note that the projectile passes through (98, 42) and (196, 0) ...

(edited 9 years ago)

Reply 10

MathMeister

OP

10

Original post by Mr M

I bet you know the equation of the trajectory.

Note that the projectile passes through (98, 42) and (0, 196) ...

You mean the y= xtan(a)-[gx^2/(2u^2)](1+tan^2(a))?

Original post by MathMeister

You mean the y= xtan(a)-[gx^2/(2u^2)](1+tan^2(a))?

Yes if you like that form.

You get a pair of simultaneous equations.

Multiply the first by 4 and subtract.

Reply 12

MathMeister

OP

10

Original post by Mr M

My head doesn't work sideways.
I wouldn't bother learning those equations. They can be derived easily whenever you need them.

You could download the pics and rotate them using windows live photo gallery or windows photo viewer.
Do you know by chance where I went wrong. (Or was it just what I did not do?)

Original post by MathMeister

You could download the pics and rotate them using windows live photo gallery or windows photo viewer.
Do you know by chance where I went wrong. (Or was it just what I did not do?)