Projectile Motion help! (2 Viewers)

[Chicken Burger]

A ball is thrown upwards at an angle of 30 degrees to the horizon with an initial velocity of 50 m/s. Calculate the:
a) height it rises
b) horizontal distance to impact
c) time to impact

Thanks

 

[Carrotsticks]

a) For maximum height, we let dy/dt=0, and you will get t = ...

Sub that t = ... back into your y = (something in terms of t) for maximum height.

b) For horizontal distance, let y=0 and similarly to above, you will get t = ...

Sub that t = ... back into x = (something in terms of t) for horizontal distance.

c) You already found time to impact by letting y=0 and solving for t in part (b).

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EDIT: Never mind this post. I did not notice that it was in the Physics section, not MX1.

 

[Chicken Burger]

Thank you math God

 

[Chicken Burger]

But can we use this method on a physics exam? Because I haven't learnt your method in physics...

 

[Secant]

wait, why is there differentiation? I don't remember differentiation being in physics projectile motion topic 1 space?

you don't do MX1 do you?

EDIT: ~peace out homie~

 

[Chicken Burger]

i do mx1 and mx2, but we havent covered simple harmonic

 

[Chicken Burger]

but we can use this method right secant? in a physics exam that is

 

[Chicken Burger]

plus i dedicated most of my time in mx1 and mx2, rarely for physics.. which is bad...

 

[Secant]

A ball is thrown upwards at an angle of 30 degrees to the horizon with an initial velocity of 50 m/s. Calculate the:
a) height it rises
b) horizontal distance to impact
c) time to impact

Thanks

a) max height: use v^2=u^2+2as (it is on the syllabus) then sub v=0, u= 50sin30 and a=-9.8 continue
b) find t using s=ut+1/2at^2 by making s=0, then use x=u(x)t, x is the range
c) u already found the time.

EDIT: ~peace out homie~

 

[jenslekman]

but we can use this method right secant? in a physics exam that is

if answer is right you will get the marks, if it is wrong (you stuffed up)you wont get marks for working

 

[Secant]

i do mx1 and mx2, but we havent covered simple harmonic

it is under 'projectile motion' in mx1 but you DO NOT need mx1 in physics and if you use mx1 you get it wrong (i think), the formulas are given to you on the syllabus and the formula sheet in the exam so just sub in and get the answer you want

EDIT: ~peace out homie~

 

[Chicken Burger]

ok thank you

 

[Chicken Burger]

a) max height: use v^2=u^2+2as (it is on the syllabus) then sub v=0, u= 50sin30 and a=-9.8 continue
b) find t using s=ut+1/2at^2 by making s=0, then use x=u(x)t, x is the range
c) u already found the time.

EDIT: ~peace out homie~

hey secant can you tell me why v=0?

 

[Chicken Burger]

bump, sorry i need to know why secant let v=0 thanks

 

[Chicken Burger]

wait, is it because at max height there is no vertical component?

 

[Frostbitten]

wait, is it because at max height there is no vertical component?

Yes, at max height the object Is only traveling horizontally.

 

[Carrotsticks]

Oh my fault! I thought this was in the Mathematics section, so I gave an answer using methods from MX1.

 

[Chicken Burger]

Oh sweet thanks frostbitten. And Carrot its fine, I learnt something new

 

[yasminee96]

i got told that using the notation in mx1 would make you lose marks. That's what i was using the whole time but apparently it makes a huge difference if you write x instead of delta x dafuq. yeh, stick to physics formulas - but they're exactly the same as mx1, just different notation, and you derive formulas in mx1

 

[Chicken Burger]

hey guys another question sorry.
A basketballer projects a ball at an angle of 30 degrees to the horizontal at 9.8 m/s.
What will be the impact velocity and its direction?