A level physics question. Help!?

Question

1 An astrounaught on the Moon threw an object 4m vertically upwards and caught it again 4.5 seconds later. Calculate:
a) the acceleration due to gravith of the Moon,
b) the spped of projection of the object,
c) how high the object would have risen on the Earth, for the same speed of projection

2 A swimmer swims 100m from 1 end of a swimming pool to the other end at a constant speed of 1.2m/s, them swims back at constant speed, returning to the starting point 210 s after starting.
Calculate how long the swimmer takes to swim from: (i) the starting end to the other end, (ii) back to the start from the other end.

Please show working out and which equation of motion used!

answers
1a 1.6m/s2
b 3.6m/s
c 0.64m

2(i) 83s
(ii)127s

Answer 1

1) a) d= (1/2)gt^2
4 = 0.5(g)(2.25 * 2.25)
g = 1.58 m/s^2

b) V = sqr rt (2 g d)
= sqr rt (2 * 1.58 * 4)
= 3.56 m/s

c) v = sqr rt (2 g d)
v^2 = 2 * 9.8 * d
d = 3.56*3.56/(2 *9.8) = 0.65 m

2) a) a = 0 (since the speed is constant)
t = d/v = 105/1.2= 87.5s

b) t = 210 - 87.5 = 122.5 s

Answer 2

If you are doing your A levels and find these questions difficult, particularly the second, maybe you should ask yourself whether A level physics is the right subject for you.

Answer 3

1]

a)a = 1.6 m/s^2 (This is a fact!)
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b)

d = vit + 1/2gt^2
4 m = vi(4.5s) + 1/2(-1.6m/s^2)(4.5s)^2
4 m = vi(4.5s) + -0.8 m/s^2(20.25s^2)
4 m = vi(4.5s) - 16.2 m
4 m + 16.2 m = vi(4.5s)
20.2 m/4.5s = vi
4.48888888888888888...m/s = vi

------We know that vf is zero------
average speed = (vf + vi) /2
average speed = (0 + 4.48888888m/s)/2
average speed = 2.2444444444444...m/s
average speed = 2 m/s (IN SIGNIFICANT FIGURES)
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C)
dy = 1/2 gt^2
dy = 1/2(1.6m/s^2)(4.5s)^2
dy = 16.2 m

Good luck and God bless!