Misc. Questions (1 Viewer)

[.ben]

Permutations

1.The number of arrangements 2n+2 different objects taken n at a time is to the number of arrangements of 2n objects taken n at a time is 14:5. Find the value of n.

Projectile Motion

2. A particle projected from a point meets the horizontal plane through the point of projection after travelling a horizontal distance a, and in the course of its trajectory attains a greatest height b above the point of projection. Find the horizontal and vertical components of the velocity in terms of a and b. Show that when it has described a horizontal distance x, it has attained a height of [4bx(a-x)]/a².

3. A batsman hits a cricket ball 'off his toes' toward a fieldsman who is 65m away. The ball reaches a maxiumum height of 4.9m and the horizontal component of its velocity is 28m/s. Find the constant speed with which the fieldsman must run forward, starting at the instant the ball is hit, in order to catch the ball at a height of 1.3m above the ground. (g=9.8)

4. Find the speed and directio of a particle which, when projected from a point 15m above the horizontal ground, just clears the top of a wall 26.25m high and 30m away.

 

[Riviet]

1. (²ⁿ⁺²P_n)/(²ⁿP_n) = 14/5

Change into factorials and solve for n. You can finish it off.

 

[.ben]

Yea, i got that one shortly after i posted

 

[pLuvia]

2.
I have the full solution but it will take too long to type all out I'll give you the gist of it.
x'=vcos@ and y'=vsin@
First let max height be b=v²/2g*sin²@ and range is a=v²/g*sin2@
Rearrange the max height so you have it in vsin@, sub that into y' and that's one answer.
Then rearrange the range "a", into vcos@, and sub this into x', and you should get your other answer.
Now that you have vsin@ and vcos@, combine both and obtain and expression for v².
Now use the equation of the path
y=xtan@-gx²(1+tan²@)/2v²

Sub in the values you have obtained, then simplify and you should get your answer
If you need anymore help just ask

 

[.ben]

Yea, i did the question but kept getting vcos@=asqrt(g/2b) instead of vcos@=(a/2)sqrt(g/2b)?
wierd.

 

[pLuvia]

Did you get vcos@=ag/2vsin@?
If so you should get the answer
You should get this expression
vcos@=ag/[2(sqrt{2gb}/sin@]*sin@

 

[.ben]

i got vcos@=asqrt(g/2b)

 

[Riviet]

Do 3 and 4 still need answering?

 

[.ben]

er. yep. for 3 i got an answer close but wrong. i got 6.2367ms/s but answer is 7m/s

for no. 4 i have no idea..

thnx

 

[.ben]

ANyone

 

[haque]

for question 4 u have to assume that the 26.25m is the maximum height-another example of fitzpatrick's dodgy expression.

 

[.ben]

yea i just found that out today thanks anyway haque.

 

[followme]

er. yep. for 3 i got an answer close but wrong. i got 6.2367ms/s but answer is 7m/s

thnx

the time of flight is 1.857 seconds, so divide distance 13 by 1.857 = 7.0005 m/s

i've edited my ans in the other thread.