Projectile Motion- Cambridge (1 Viewer)

[kaz1]

I can't seem to be able to do b)i. I keep getting cos^-1 3/4. Help would be appreciated. The question is Q16 on page 131 in Year 12 Cambridge 3unit.

 

[shaon0]

I can't seem to be able to do b)i. I keep getting cos^-1 3/4. Help would be appreciated. The question is Q16 on page 131 in Year 12 Cambridge 3unit.

hey Kazi. Here's the solution:

16a) When y' = 0.
vsin@ = gt
t = (vsin@)/g .....sub into y=(-g/2)t^2 + vtsin@

y=(-g/2)( (vsin@))^2 + v( (vsin@))sin@
y=(v^2(sin(@))^2)/2g

b) i) y{P2} = (3/2V)^2(sin(@/2))^2/2g
= (9V^2)(sin(@/2)^2/8g

For y = (v^2sin(@/2+@/2))/2g
= 4V^2 (sin(@/2))^2((cos(@/2))^2/2g
Let y{P2} = y

end up getting:
9/16 = (cos(@/2))^2
@/2 = arccos(3/4)
@ = 2arccos(3/4)
= arccos(1/8)

ii) Yes because there's only one external force affecting the fight path of the projectile ie gravity. And since the have the same initial conditions.