am 1998 Q.1(a)

Shox

Just can't seem to see the little trick in it???

Shox

And for 1986 Q.3 the last part,

To calculate t1 isnt it

Tan A = -Vy/vx
4 = -Vy/Vx

Only i got 8 for the answer instead a 4.25???

cocoa

T= total time
x=time taken for acceleration
y=time at full speed
z=time taken for deceleration

therefore the fraction of journey time spent at full speed, what we want, is y/(x+y+z) or y/T

(5v/6)*T = xv/2 + zv/2 + yv
cancle v
5T/6 = (x+z)/2 + y
5T/6 = (T-y)/2 + y
5T/6 = T/2 -y/2 +y
2T/6 = y/2
4/6=y/T
y/T=2/3 ANS

and for your second question...
I wouldn't really expect the answer to be 8s, considering that's the moment it hits the horizontal...

ok, so at t=0
u(up)=sin(alpha)u
u(across)=cos(alpha)u
a=g downward

finally, at t=8
s(up)=0=ut+at^2/2=u+at/2 [we don't want the instance of t=0, we already know s=0 at that point] 0=u sin(alpha) - gt/2

u=4g/sin(alpha) [but we know from the first part that tan(alpha)=4]
u=groot(17)

at t=t1
u(up)= -cos(alpha)x
u(across)= sin(alpha)x

u(across) hasn't changed, therefore sin(alpha)x = cos(alpha)u
x=u/tan(alpha)
u(up)=-cos(alpha)u/tan(alpha)
and, of course, u(up) is also = u + at = sin(alpha)u - gt1 = -cos(alpha)u/tan(alpha)
[ok, so now i'll finally remove alpha... sorry i didn't do it sooner to be honest...]
4u/root(17) -gt1 = - u/4root(17)
[and i'll remove u as well...]
4groot(17)/root(17) -gt1 = -groot(17)/4root(17)
4g - gt1 = -g/4
4 - t1 = -1/4
t1 = 4 + 1/4 = 4.25 ANS

sorry if that's a bit long winded and hard to read, hopefully you can find your mistake.