A ball is projected vertically upwards.Two light gates are placed along its route at different heights as shown.The timer-scaler records that the ball first passes light gate A and light
gate B 0.200 s and 0.458 s after the projection respectively.The height of light gate A and light gate B are 2 m and 4 m respectively.

(a)Use the above information to estimate the projection velocity u and the accleration due to gravity g.
ans:u=11.0 m/s (upwards),9.82 m/s(downwards)

(b)Calculate the times that the ball passes the light gates again.
ans:1.78 s,0.458 s

(c)What is the velocity of the ball when it passes light gate A
(i)for the first time?
ans:9.02 m/s
(ii)for the second time?
ans:9.02 m/s

謝解答,thx

a.
s=ut+1/2at^2
4=u(0.458)-1/2g(0.458)^2
4=0.458u-0.105g---1.
---------------------------------------
s=ut+1/2at^2
2=u(0.2)-1/2g(0.2)^2
2=0.2u-0.02g---2.
By sub數就會搵到答案

b.(應該係搵B?)
s=ut+1/2at^2
4=11t+1/2(-9.82)t^2
4=11t-4.91t^2
4.91t^2-11t+4=0
t=1.78s or 0.457s

c.
(i.)(ii.)
v^2=u^2+2as
v^2=11^2+2(-9.82)(2)
v^2=81.72
v=+-9.04ms^-1
上係+,落係-