more help with projectile motion (1 Viewer)

[jemsta]

geez i need help with projectile motion,
1) find an expression for the maximum range in the horizontal plane for a particle with initial velocity Vm/s. find the two angles of projection to give a range equal to half the maximum range.
2) a projectile has a trajectory such that its range on a horizontal plane is four times the maximum height attained during flight. find the angle of projection

 

[klaw]

1)
When y=0, Vtsin@-gt²/2=0
t=2Vsin@/g

x=Vtcos@
When t=2Vsin@/g
x=2V²sin@cos@/g
=V²sin2@/g

When range=max, @=45
.:Max range =V²/g

When range = V²/(2g),
V²/(2g)=V²sin2@/g
1/2=sin2@
2@=pi/6, 5pi/6
.=pi/12, 5pi/12

 

[klaw]

When y'=0, Vsin@-gt=0
t=Vsin@/g

When t=Vsin@/g,
y=V²sin²@/2g

.: Max Height = V²sin²@/2g
Horizontal range=V²sin2@/g (from Q1)

When horizontal range is 4* max height,
8V²sin2@=Vsin²@
8sin2@=sin²@
16sin@cos@=sin²@
sin@(sin@-16cos@)=0
.: sin@=0 (rejected) or sin@=16cos@
tan@=16
.=tan^-116