Projectile Motion Help (1 Viewer)

[Kimyia]

We got this question in class: a projectile is fired with an initial velocity of 1000m/s with an angle of 30 with the ground. Find the maximum height, the time to reach the maximum height, and the range.
I found the maximum height ok: 12755m
But I got a completely different answer for the time to reach the maximum height. Could someone please show me how you would work it out?
Thanks!

 

[Carrotsticks]

Did you do the following steps:

Sub max height into y (as a function of t)

Then solve for t?

 

[Kimyia]

I used the delta y = uyt + 1/2ayt^2 using the max height of y but they used vy = uy + ayt

 

[Carrotsticks]

Oh dear this is the Physics forum not the Maths Extension 1 forum.

Sorry, I don't know what the Physics formula is or how to use it.

 

[Timske]

We got this question in class: a projectile is fired with an initial velocity of 1000m/s with an angle of 30 with the ground. Find the maximum height, the time to reach the maximum height, and the range.
I found the maximum height ok: 12755m
But I got a completely different answer for the time to reach the maximum height. Could someone please show me how you would work it out?
Thanks!

Max height
u = 1000m/s
angle = 30
uy = 1000sin30 = 500m/s
ux = 1000cos30 = 866m/s

uy = 500m/s vy = 0m/s ay = -9.8m/s^2 y= ?
v^2 = u^2 + 2ay
0 = 500^2 + 2*-9.8*y
0 = 250000 -19.6y
19.6y = 250000

Trip time
uy = 500m/s vy= 0m/s ay=-9.8m/s^2 t= ?
v = u + at
0 = 500 -9.8t
9.8t = 500
t = 51s
NOTE: this is the time taken to rise to the peak. If the question asked for total time of flight you times 2.

Range
ux = 866m/s trip time = 102s range?
delta x = uxt
= 866*102
= 88332m

 

[Kimyia]

Trip time[/B]
uy = 500m/s vy= 0m/s ay=-9.8m/s^2 t= ?
v = u + at
0 = 500 -9.8t
9.8t = 500
t = 51s
NOTE: this is the time taken to rise to the peak. If the question asked for total time of flight you times 2.

Thanks for the help. My question now is, why do we use vy = uy + ayt and why can't we use delta y = uyt + 1/2ayt^2?

 

[qrpw]

You can. It should give you the same answer. Only problem is it'll give you an annoying quadratic in t (easily solvable, but you'd rather use an easier method).

 

[nightweaver066]

Thanks for the help. My question now is, why do we use vy = uy + ayt and why can't we use delta y = uyt = 1/2ayt^2?

Misread it. Yeah you can use it, will give you the same answer and you just disregard t = 0 because that's when it is initially launched.

 

[SpiralFlex]

Did you do the following steps:

Sub max height into y (as a function of t)

Then solve for t?

The physics formula is the same as the ones you use for 3U projectiles, however rewritten in fancy subscripts to accommodate the needs of a Physics student who may not do mathematics.

Someone, please start a Physics Marathon soon.

 

[nightweaver066]

The physics formula is the same as the ones you use for 3U projectiles, however rewritten in fancy subscripts to accommodate the needs of a Physics student who may not do mathematics.

Someone, please start a Physics Marathon soon.

Why don't you Spiral?

 

[SpiralFlex]

Why don't you Spiral?

No! I already got a marathon for SDD *(tumbleweeded)*. I decorated with pretty colours. But okay. I guess, sometime this week then...

 

[Timske]

Why don't you Spiral?

his busy trying to save his gf

 

[Kimyia]

I've got it now, thanks everyone!

 

[turbo2007]

Nice Job