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Hi Paul,

A projectile is fired at an initial speed of 500m/s and angle of
elevation of 30 degrees. Find:
(a)   Range of the projectile
(b)   Maximum height reached
(c)   The speed at impact.

(Working out included)



X=500 cos 30t= 433t
Y=(500 sin 30)t-0.5gt^2
= (250)T-4.9T^2

By plotting the maximum occurs at point (25.51,3188.775)

So 3188m is the maximum height.

For the speed at impact I  went:

Y=(500 sin 30)t-0.5gt^2
= (250)T-4.9T^2

Impact occurs when y=0 so t=51 metres. So v(t) = r[?][?][?](t)=xi+yj

So: 433i + 250T-4.9T^2j

Since t = 51 then 51 * 433 =22083metres hits the ground at this range

433i + (433 [?][?][?] 9.8T)J

V(51) = (433)^2 + (433-9.8*22083)^2

Also, How do I find the speed upon impact? I tried using the  derivative
of r(t). After that I plugged v(51) which is the derivative of r(t)I
arrived at a large figure which is impossible in a common sense way. I
believe I am on using the correct idea on how to solve this but it seems

wrong because the figures are unreasonable I must have made a mistake somewhere at the start of this problem.

Regards,

Lilin Chen

Answer
Questioner:   Lilin Chen
Country:   Western Australia, Australia
Category:   Advanced Math
Private:   No
Subject:   Calculus Kinematics
Question:   Hi Paul,

A projectile is fired at an initial speed of 500m/s and angle of
elevation of 30 degrees. Find:
(a)   Range of the projectile
(b)   Maximum height reached
(c)   The speed at impact.

I hate doing things with calculators.  So I did this all on the assumption that

Initial velocity = v0, at an angle of 30 degrees.

x-component written v0x = v0 cos 30  <<<<<<<<<<< s3  means square root of 3
y-component is v0y = v0 sin 30

cos 30 = s3/2
sin 30 = 1/2

.................................
So v0 = 500, v0y = v0 sin 30 = v0/2

And y0 = 0.

y = -16t^2 + v0/2 t       <<<< basic height equation

v = dy/dt = -32t + v0/2

Set = 0,  -32t + v0/2 = 0

tmax = v0/64

At tmax, y = -16(v0/64)^2 + v0/2 (v0/64)

= -16(v0/64)(v0/64) + v0/2 (v0/64)

= v0^2 [- 1/256 + 1/128]

= v0^2 [1/256] = v0^2/128  <<<< answer to (b)
....................

At impact,  tcrash = 2*tmax = v0/32

Now we look at x's:

v0x = v0 cos 30 = s3/2 <<<<<<<<<<< s3  means square root of 3

x-distance traveled = v0x * tcrash

= v0 cos30 * v0/32

= v0^2 (s3/64)   <<<<<<< that is the range; answer to (a)

........................

At impact,  tcrash = 2*tmax = v0/32

vx = v0 cos 30 = v0 * s3/2   <<<<< vx never changes

vy = -32t + v0/2

vycrash = -32(tcrash) + v0/2

= -32(v0/32) + v0/2

= - v0 + v0/2 = - v0/2   <<<<  duh!!!!! (or the Aussie equivalent) The downward velocity is exactly the opposite of the upward!

So the impact speed (absolute value) is exactly the initial speed.  (You could have figured that out, right?)

NOW you can plug in your v0 = 500 and start using your calculator.

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