Projectile motion

a)個高度係受地心吸力影響..而唔關horizontally的力事..
假設冇空氣阻力..
u=0 v=?? a=10(take下為正) t=1.5 s=??
s=ut+(1/2)at^2
s=0+10x1.5^2/2
s=11.25m
so position 50-11.25=38.75m

b)
u=0 v=?? a=10 t=?? s=50
s=ut+(1/2)at^2
50=0+5t^2
t^2=10
t=3.162277

c)
u=200 v=0 a=?? t=3.16227 s=??
s=200x3.16227
s=632.454m

d)我唔知呢個係計垂直的velocity還是垂直加水平的velocity
垂直的velocity:
u=0 v=?? a=10 t=3.16227 s=50
v=u+at
v=10x3.16227
v=31.6227m/s

如果係垂直加水平的velocity
31.6227^2+200^2=v^2
v^2=41000
v=202.485m/s

1 since no net force acting on the bullet horizontally, its horizontal velocity maintain 200m/s. only mg acting on bullet vertically, so,(By mg=ma, g=a) there is a constant acceleration on its vertical velocity.
to solve this problem, we need to consider both vertical(V) and horizontal(H) case
(a)take downwards as positive
we need consider both distance of V and H
for H, 200x1.5=300m
for V, u=0, v=?, t=1.5 ,a=10, s=ut+(1/2)at^2
s=0x1.5+(1/2)x10x(1.5)^2
s=11.25m
after 1.5s, the bullet is 11.25m of vertical and 300m of horizontal from its origin point
(b) all we need to know is when the bullet hit the ground(垂直地走完50m), so, just considerV
s=50, a=10,u=0 t=?
50=0Xt+(1/2)10xt^2
t^2=10
t=開方10=3.16s
(c) in this case , we only consider H( tower的底至到子彈到地的距離)
when bullet hit the ground,it is 3.16s
so, 200x3.16=632m
(when bullet hit the ground, it has travelled 632m of H)
(d) velocity=speed +direction
for speed of H, 200m/s
for speed of V, u=0,v=?,t=3.16,a=10
v=u+at
v=0+10x3.16=31.6m/s
這是垂直和水平的速，但我地平時見物體落
所以我地要計打的速, 用畢氏定理
v^2=(31.6)^2+(200)^2
v^2=40998.56, v=202m/s
direction： 31.6/200=tanO
O=8.98*
the bullet moves with 202m/s and 8.98度 with the horizontal
我英文不好，請自行理解上文