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HSC Stuff => HSC Science Stuff => HSC Subjects + Help => HSC Physics => Topic started by: BakerDad12 on October 28, 2019, 07:37:30 pm

Title: Projectile motion question
Post by: BakerDad12 on October 28, 2019, 07:37:30 pm

I cannot do this projectile motion question:

A Bofors gun that releases a shell at an initial velocity of 880m/s launches a shell at angle of 60 degrees. At what time and height does the shell have a vertical velocity of zero?

Thank you!!

Title: Re: Projectile motion question
Post by: harry.braithwaite on October 28, 2019, 08:54:43 pm

First you want to find the height, then you can find the time to reach that height. (Apologies, haven't learnt how to use LaTex yet)

So we start with v^2 = u^2 + 2as (vertically), and want to find when v = 0, so with u = 880sin60 m/s, a = -9.8 m/s^2, and s = ? (setting up as the positive direction

Rearranging we get s = -(u^2)/2a, plugging in the above values we get s = 29,632.65... m, which is our max height

Now plugging this into s = ut + 1/2a(t^2), and above values, we use quadratic formula to rearrange and find that t = (-u-((u^2)+2*a*s)^0.5)/(a), which gives a time til max height as 77.77 seconds (2dp) (remembering that a = -9.8 )

Hope this helps

Title: Re: Projectile motion question
Post by: BakerDad12 on October 30, 2019, 07:28:38 am

Right, thank you!