力學！mechanics

Question

1)A bullet is fired towards a nearby tree with a speed of 200ms^-1 .the bullet is later found at a depth of 5cm.Find the average deceleration of the bullet inside the tree.
2)A ball is released from rest and falls vertically under gravity.If the distance travelled by the ball in the 1st second is p and that travelled in the 2nd second is q,what is the ratio of q:p?
唔該寫得詳細一點！thx~

Answer 1

1. u=200 m/s, v=0 m/s, s=5/100 m, a=?

using the equation of motion: v^2 = u^2 +2.a.s
0 = 200^2 + 2.a.(5/100)
this gives a = -400 000 m/s2 (the -ve sign indicates a decceleration)

2. For the first second,
u=0 m/s, a=g (acceleration due to gravity), t= 1s, s=p

using s = ut + (1/2)a.t^2
p = (1/2).g(1)
this gives 2p = g -------------- (1)

For the first TWO seconds,
u = 0 m/s, a=g, t =2 s, s = p+q
using again s = ut + (1/2)a/t^2
p+q = (1/2)g(2x2)
this gives p+q = 2g ------------- (2)

Dividing (2)/(1): (p+q)/2p = 2g/g
p+q = 4p
thus q = 3p
i.e. q/p = 3