tohero:

Think about you throwing me a stone at a fixed distance.
If you want the stone to hit me hard, you must throw it directly at me. If you threw it in a parabolic form in the sense that it went up and on coming down it hit me, believe me, it is not going to be painful as the previous.
NB:
-This distance between us doesn't determine how painful it will be to me.
-what determines it is the height of where I am which will prompt the angle of elevation.
I hope you get the analogy.

well, I think for each hit, there are two firing angles (as long as the firing angle isnt 45degrees) namely theta and 90-theta. I also think the question was asking for the lesser of the two since the greater of the two will make the particle spend longer time in the air.

tohero:


Using the formula v² = u²+2gh
Where
v=0 ,
g=-9.8m/s²
h= 50 - 1.8 = 48.2m
Vertical component of the initial velocity = u sinA
Where A=75^o.

Then the formula bcoms
0² = u²sin²75^o - 2*9.8*48.2

That is
19.6*48.2 = u²sin²75^o

Calculator should finish this.

why did you set v=0? Are you assuming the height to be the maximum height? It wasn't given like that nah. What do you think?

benji93:


the catapult was held at 1.5m above the ground means it is 28.5m below the bird vertically.
the poacher is 25m from the tree horizontally
we have a right angled triangle with 28.5m as the opposite and 25m as the adjascent, and the path taken by the catapult is the hypotenuse
x^2 = 25^2 + 28.5^2
x^2 =1437.25
x=37.911m
duration of shot = 37.911/40=0.955s
tan theta = 28.5/25
theta = tan inverse of 28.5/25=0.851m

are you using pythagora's? I think it would work like that if there's no acceleration due to gravity acting towards the surface of the earth

jackpot:

why did you set v=0? Are you assuming the height to be the maximum height? It wasn't given like that nah. What do you think?

We are looking for the initial velocity which means the object has been stationary, hence, it implies that velocity=zero.

jackpot:
well, I think for each hit, there are two firing angles (as long as the firing angle isnt 45degrees) namely theta and 90-theta. I also think the question was asking for the lesser of the two since the greater of the two will make the particle spend longer time in the air.

We are already given the angle as 75^o. What other angles are you suggesting for
Our aim is to find the initial velocity not --the lesser of two angles.

That's my solution anyway. Bring forth your possible solution probably we may get convinced.

jackpot:
are you using pythagora's? I think it would work like that if there's no acceleration due to gravity acting towards the surface of the earth

rit if gravity is taken into consideration then s= ut + (1/2)gt^2,but since it is shot at a component of the velocity would be acting vertically upwards
so we can find theta first using the sides of the right angled triangle and then find t since we know u,g and we have found s=x=the hypotenuse.
or probably s = usin theta + (1/2)gt^2,where s is the vertical of the right angled triangle,

jackpot:
Question four



A poacher standing at a distance of 25m from a tree saw, aims and takes a shot with a catapult at a bird whose vertical distance from the ground is 30m. Suppose that
(i) the catapult is held at a distance of 1.5m above ground level,
(ii) the initial velocity of the shot was 40m/s,
(iii) the bird was hit before it could make a move,
then calculate
(a) the duration of the shot
(b) the angle of the shot

it would be easier to solve all this ur projectile if u consider just the vertical distance, horizontal & tan of the angle at first. They are cases of projectiles of say a footballer hitting a ball at angle theta & the ball landing on a roof ymetres from the ground at a distance xmetres from the footballer.

If you do not follow the approach I gave, you will probably bask in the euphoria that the answer you got was correct when it isn't.

Miscellaneous:


it would be easier to solve all this ur projectile if u consider just the vertical distance, horizontal & tan of the angle at first. They are cases of projectiles of say a footballer hitting a ball at angle theta & the ball landing on a roof ymetres from the ground at a distance xmetres from the footballer.

If you do not follow the approach I gave, you will probably bask in the euphoria that the answer you got was correct when it isn't.

alright, Sir. solve one please.

tohero:



We are looking for the initial velocity which means the object has been stationary, hence, it implies that velocity=zero.


We are already given the angle as 75^o. What other angles are you suggesting for
Our aim is to find the initial velocity not --the lesser of two angles.

That's my solution anyway. Bring forth your possible solution probably we may get convinced.

thanks, it was a mix-up. I was referring to a similar question in my text.

jackpot:
alright, Sir. solve one please.

lol………

I'm relaxing & I'm not with calc. here

use, y= (xtan©) - { (gx²)/(2u²cos²©)}

where;
y=vertical distance
x=horizontal distance
© = given angle
u= initial velocity


also put R= Ux • T into use

R= range
Ux = ucos©
T= flight time

also put say, Vy = usin© - gT into use;
Vy = vertical component of velocity

try it ……… u na scholar na!

jackpot:
are you using pythagora's? I think it would work like that if there's no acceleration due to gravity acting towards the surface of the earth

angle = 48.743

hmmmm.

see math scholars abeg .

busy solving with passion , greets you all guys



@ jackpot sorry couldn't post solutions, but am glad. guys are already doing so ,

really gat tight schedules with school tinz & other engagements .



Nice solving

#shalom

Miscellaneous:


lol………

I'm relaxing & I'm not with calc. here

use, y= (xtan©) - { (gx²)/(2u²cos²©)}

where;
y=vertical distance
x=horizontal distance
© = given angle
u= initial velocity


also put R= Ux • T into use

R= range
Ux = ucos©
T= flight time

also put say, Vy = usin© - gT into use;
Vy = vertical component of velocity

try it ……… u na scholar na!

lol. You too na super-scholar. Take five!

jackpot:
lol. You too na super-scholar. Take five!

lol...........ur own na just to make fun of guyz here.......loool

Help me with these questions ; find the values of x and y in ; x^y+y^x=17 and x+y=5. (2) find the value of x in 4^x=8x.

Richiez:
We diagnose and solve math problems here
This thread is the meeting point for nairaland math gurus...I dare anyone to ask a question in mathematics without me having an answer to them, LETS START

Help me with these questions ; find the values of x and y in ; x^y+y^x=17 and x+y=5. (2) find the value of x in 4^x=8x.

hi

hi

benji93:

angle = 48.743

show steps, Sir

jackpot:
show steps, Sir

tan theta = 28.5/25
theta = 48.743degrees

benji93:

tan theta = 28.5/25
theta = 48.743degrees

you're using SOHCAHTOA? _{***covers face***}

Differentiate X^x^x

Nwiboazubuike:
Differentiate X^x^x

x^x^x[ x^x(lnx+1)lnx + x^(x-1) ]



don't ask 4 workings except u wanna learn not test our intelligence.

Nwiboazubuike:
Differentiate X^x^x

Ok, here is the solution.
Let Y = X^x^x
firstly... let U = x^x so that... Y = X^U
Now... dy/dx will be equal to dY/dU x dU/dx
Considering U = x^x first.
Applying natural logarithm to the base of exponential to both sides... we have
lnU = lnx^x which can be written as lnU = xlnx,
Now, the derivative of the equation gives
(1/U)dU/dx = x(1/x) + lnx(1)... Using product rule. And don't forget that, U is differentiated implicitly with respected to x.
(1/U)dU/dx = 1 + lnx
Therefore dU/dx = U(1 + lnx).
Since U = x^x, therefore dU/dx = x^x(1 + lnx).

Alright, considering Y = X^U also.
Applying log into base of e to both sides... we have
lnY = UlnX... Also, following the same rule for this also... we have
(1/Y)dY/dU = U(1/x) + lnx(x^x + x^xlnx)... since du/dx = x^x + x^xlnx
(1/Y)dY/dU = x^x(1/x) + x^xlnx + x^x(lnx^2)
dY/dU = x^(x+1) + x^xlnx + x^x(lnx^2)

So therefore dY/dx = dY/dU x dU/dx

which is equal to x^(x+1) + x^xlnx + x^x(lnx^2) x x^x(1 + lnx).

Finally dY/dx = [x^(x+1) + x^xlnx + x^x(lnx^2)] X [x^x(1 + lnx)].

You can expand further by opening the bracket.

@ factorial1 u try sir ,

differential calculus is quite cheep , but even though integral calculus is the reverse of it , its not usually funny to do so . Now in this regards , how can we then obtain back the integrand .?

agentofchange1:
@ factorial1 u try sir ,

differential calculus is quite cheep , but even though integral calculus is the reverse of it , its not usually funny to do so . Now in this regards , how can we then obtain back the integrand .?

Lol... By finding the integral of course, which I'm sure won't be quite easy.

Nature130:
Help me with these questions ; find the values of x and y in ; x^y+y^x=17 and x+y=5. (2) find the value of x in 4^x=8x.

there exist NO analytical solutions to such equations yet , except approximate or graphical

though

1) (x,y) =(2,3)=(3,2)

2) x=2

you could be the first mathematician to develop that ,

happy trying .

factorial1:
Lol... By finding the integral of course, which I'm sure won't be quite easy.

sure , that's the point .

Nature130:
Help me with these questions ; find the values of x and y in ; x^y+y^x=17 and x+y=5. (2) find the value of x in 4^x=8x.

May not be able to type the solutions tonight but lemme give an Hint to solving question 2.
Since 4^x = 8x... Divide both sides by 4 to get 4^(x-1) = 2x.
Don't forget 4^(x-1) can be written as (1 + 3)^(x-1)... equating that to 2x gives
(1 + 3)^(x-1) = 2x... Then solve using Binomial Expansion . The answer should be 2 and 0 cause it will surely lead to quadratic equation. Note that you only need the first 3 terms of the expansion.

agentofchange1:



there exist NO analytical solutions to such equations yet , except approximate or graphical

though

1) (x,y) =(2,3)=(3,2)

2) x=2

you could be the first mathematician to develop that ,

happy trying .

Question 2 is solvable bro.. Try it out. Gonna lead to you using Binomial expansion.
Can't really be typing tonight... Have got a lot.

factorial1:
May not be able to type the solutions tonight but lemme give an Hint to solving question 2.
Since 4^x = 8x... Divide both sides by 4 to get 4^(x-1) = 2x.
Don't forget 4^(x-1) can be written as (1 + 3)^(x-1)... equating that to 2x gives
(1 + 3)^(x-1) = 2x... Then solve using Binomial Expansion . The answer should be 2 and 0 cause it will surely lead to quadratic equation. Note that you only need the first 3 terms of the expansion.

nice try sir, av known of this method too but the problem is , its still an approximate, since we truncate @ the 3rd term of expansion , why not 4th or 5th or higher ?. that's because we already know the answer to be 2 by trial -by-error , which is not always the ideal way of solving mathematical problems .


Now can it work for this ?

4^x = x^4 .?

or 3x = 8^x

agentofchange1:


nice try sir, av known of this method too but the problem is , its still an approximate, since we truncate @ the 3rd term of expansion , why not 4th or 5th or higher ?. that's because we already know the answer to be 2 by trial -by-error , which is not always the ideal way of solving mathematical problems .


Now can it work for this ?

4^x = x^4 .?

or 3x = 8^x

Solving that question doesn't require using all the terms...as a matter of fact... not for only this question. There are some maths question that one will just have to make use of some part of the equation.