Fitzpatrick Projectile Motion Question Help... (1 Viewer)

[Shikobe]

25D, Question17.

A stone is throne so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed of 10m/s. The stone reaches a height double that of the pole and, in its descent, touches the bird. Find the horizontal component of the velocity of the stone.

There are solutions on bored of studies but i dont understand them so if anyone could help me i would greatly appreciate it!

 

[hscishard]

25D, Question17.

A stone is throne so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed of 10m/s. The stone reaches a height double that of the pole and, in its descent, touches the bird. Find the horizontal component of the velocity of the stone.

There are solutions on bored of studies but i dont understand them so if anyone could help me i would greatly appreciate it!

Thrown. Lol

 

[Ferox]

(I'll use ' instead of the usual dots)<o></o>
<o></o>
x''=0 (1a)<o></o>
x'=Vcosθ (1b)<o></o>
x=Vtcosθ (1c)<o></o>
<o></o>
and<o></o>
<o></o>
y''=-g (2a)<o></o>
y'=Vsinθ -gt (2b)<o></o>
y=Vtsinθ -½gt²(2c)<o></o>

Find the max height<o></o>
y'=0 => t=(V/g)sinθ<o></o>
-->(2c)=> y=(V²/g)sin²θ - (V²sin²θ)/(2g) = (V²sin²θ)/2g
<o></o>
at y=2h<o></o>
<o></o>
2h = (V²sin²θ)/2g => Vsinθ = 2√(gh)<o></o>
<o></o>
Now, consider when y=h<o></o>
<o></o>
h = Vtsinθ -½gt² <o></o>
½gt² (2c) - 2√(gh) + h =0<o></o>
<o></o>
t= [2√(gh) ± √(4gh - 4(½g)(h))] / g<o></o>
=[2√(gh) ± √(2gh)] / g<o></o>
<o></o>
That's the times the rock's at h height. <o></o>
<o></o>
Now the bird travels for the longer time, i.e.<o></o>
<o></o>
Time of Bird's flight = [2√(gh) + √(2gh)] / g<o></o>
<o></o>
and he travels at 10m/s Therefore the distance is that time times 10, i.e.<o></o>
D= 10[2√(gh) + √(2gh)] / g<o></o>
<o></o>
But the time it takes the rock to travel that distance is the longer time - the shorter time, i.e.<o></o>
<o>
</o> T = [2√(gh) + √(2gh)] / g - [2√(gh) - √(2gh)] / g
= (2/g)√(2gh)<o></o>
<o></o>
and remembering speed=distance/time<o></o>
vcosθ = 10[2√(gh) + √(2gh)] / g * g/[2√(2gh)]<o></o>
=(5/√ 2)(2+√2)<o></o><o>
</o>

 

[Shikobe]

ThankYou!!!