t = 0.19s u_{V}= 0

s_{V}= ut + 1/2 at^{2}= 0 + 1/2 x^{-}10 x 0.19^{2}= 0.18 m (18 cm)

u_{H}= 35 cos 20° = 32.9 ms^{-1 }u_{V}= 35 sin 20° =^{+}12.0 ms^{-1 }s_{V}= - 50 m

(a) v_{V}^{2}= u_{V}^{2}+ 2as = (^{+}12)^{2}+ 2 x^{-}10 x^{-}50

v_{V}^{ }= ± 33.8 ms^{-1}(must be negative, hence v_{V}= - 33.8 ms^{-1})

v_{V}= u_{V}+ at

- 33.8 =^{+}12^{ }+^{-}10 t

t = 4.58 s

Using trigonometry: v (the hypotenuse) = 47.2 ms^{-1 }The angle q = tan^{-1}(33.8/32.9) = 45.8° (to horizontal).

(c) s_{H}= v_{H}t = 32.9 x 4.58 = 150.7 m

Maximum range: R = u^{2}/a = 9.228 m

He was 0.328 m off maximum (96.5%).

The diagram showing the 53° angle is as follows:

The trigonometric relationship between the velocities is as follows:

u_{V}= v cos q ; u_{H}= v sin q

(a) s_{V}= u_{V}t + 1/2 at^{2 }-730 = u_{V}x 4.5 + 1/2 x^{-}10 x 4.5^{2 }-730 = 4.5 u_{V}+^{-}101

u_{V}=^{-}140 ms^{-1 }u_{V}also equal v cos 53° so:^{-}140 = v cos 53°

v = -140/0.60 =^{-}233 ms^{-1 }at an angle of 53° downward

(b) s_{H}= v_{H}t = v sin 53° x 4.5

= 233 x 0.8 x 4.5 + 186.4 x 4.5 = 837 m

(c) v_{V}= u_{V}+ at =^{-}140 +^{-}10^{ }x 4.5 =^{-}185 ms^{-1 }Using trigonometry:

v = 263 ms^{-1}(hypotenuse); q = tan^{-1}(185/186.4) = 44.8° to the horizontal

(a) F_{f }= m F_{N}so for a person to remain there, the friction must equal their weight (F_{W}):

F_{f}= m F_{N}= F_{W}= mg

The normal force F_{N}is equivalent to the centripetal force F_{C}which equals mv^{2}/r.

Hence:

m F_{N}= mg

m F_{C}^{ }= mg

m mv^{2}/r = mg

m v^{2}/r = g

v^{2}= gr/m = 10 x 2.1/0.4 = 52.5

v = 7.2 ms^{-1}

(b)

F_{C}= mv^{2}/r = 49 x (7.2)^{2}/ 2.1 = 1225 N

(a) a = v^{2}/r = 120^{2}/800 = 18 ms^{-2}(=1.8 'g').

(b) 1.8 'g' < 4.5 'g' so he will remain conscious.

(c) T = F_{C}+ F_{W}= ma_{c}+ mg = 80 x 18 + 80 x 10 = 2240 N

(a)

(i)

(ii)

(iii)

(iv)

(b)

(i) Same as (i) above except that the displacement gets less with each cycle.

(ii) Same as (ii) above except that the velocity gets less with each cycle.

(iii) Same as (iii) above except that the acceleration gets less with each cycle.

(iv) Spiral in towards the centre.

(c) The graph seems to be attracted to something at the origin of the two axes.

(a) s_{V}= u_{V}t + 1/2 at^{2 }-20 = 0 + 1/2 x^{-}10^{ }t^{2 }t = 2 s

s_{H}= v_{H}t = 145000/3600 x 2 = 80 m

(b) q = tan^{-1}80/20 = 76°

(c) v_{V}= u_{V}+ at = 0 +^{-}10 x 2 =^{-}20 ms^{-1 }v_{H}= 40.3 ms^{-1 }vector diagram: <img src = "ch5q50b.gif">

Impact

By Pythagoras' Theorem: v = 44.7° ; q = tan^{-1}20/40 = 27°

velocity: v = 80% x 44.7 = 35.8 ms

Launch^{-1}; q = 27°

u_{V}= v sinq = 35.8 sin 27° = 16 ms^{-1 }is when v

Maximum height_{V}= 0:

v_{V}^{2}= u_{V}^{2}+ 2as

0 = 16^{2}+ 2 x^{-}10 x s

s_{V}= 12.8 m (maximum height at next bounce)

= u

Range^{2}/a x sin 2q = 35.8^{2}/10 x sin (2 x 26.6° ) = 102 m

Alternativelyby calculating time of flight:

v_{V}= u_{V}+ at

-16 = +16 +^{-}10 t

t = 3.2 s (time of flight)

s_{H}= v_{H}t = v cos 26.6° x 3.2 = 35.8 x cos 26.6° x 3.2 = 102 m