BdPhO Regional (Dhaka-South) Secondary 2019/7

Discuss Physics and Physics Olympiad related problems here
User avatar
SINAN EXPERT
Posts: 38
Joined: Sat Jan 19, 2019 3:35 pm
Location: Dhaka, Bangladesh
Contact:

BdPhO Regional (Dhaka-South) Secondary 2019/7

Unread post by SINAN EXPERT » Mon Jan 28, 2019 9:53 pm

BdPhO Regional (Dhaka-South) Secondary 2019(7).png
BdPhO Regional (Dhaka-South) Secondary 2019(7).png (26.72 KiB) Viewed 400 times
Adventurous Azmain slides down a water slide. He starts with zero initial velocity. The initial point $A$ is at a height $H$ from the ground and the other end $B$ is at a height $h$. The tangent line to the water slide at the point $B$ makes an angle $θ$ with horizontal.
$C$ is the point on ground straight down to point $B$.
How far from the point $C$ does Azmain fall on the ground?
$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$

User avatar
SINAN EXPERT
Posts: 38
Joined: Sat Jan 19, 2019 3:35 pm
Location: Dhaka, Bangladesh
Contact:

Re: BdPhO Regional (Dhaka-South) Secondary 2019/7

Unread post by SINAN EXPERT » Wed Apr 24, 2019 8:29 pm

Let the point where the water slide touches the ground be $D$. At point $D$, $v_0^2=2gH$.
At point $B$, $v^2=v_0^2-2gh=2g(H-h)$.
Now, Azmain should travel on PROJECTILE MOTION.(If you have no idea about projectile motion, you can click on the link for easy explanation.)
Velocity among $X$ axis, $v_x=vcosθ$.
Velocity among $Y$ axis, $v_y=vsinθ$.
Now, total time required, $t=t_1+t_2=\dfrac{2vsinθ}{g}+\dfrac{\sqrt{v^2sin^2θ+2gh}-vsinθ}{g}=\dfrac{vsinθ+\sqrt{v^2sin^2θ+2gh}}{g}$
So, distance from point $C$,

$d=v_xt=vcosθ\dfrac{vsinθ+\sqrt{v^2sin^2θ+2gh}}{g}=\sqrt{2g(H-h)}cosθ\dfrac{\sqrt{2g(H-h)}sinθ+\sqrt{2g(H-h)sin^2θ+2gh}}{g}=\dfrac{2g(H-h)cosθ sinθ+\sqrt{4g^2(H-h)^2sin^2θ cos^2θ+4g^2(H-h)h cos^2θ}}{g}=\dfrac{2g(H-h)cosθ sinθ+2g\sqrt{(H-h)^2sin^2θ cos^2θ+(H-h)h cos^2θ}}{g}=2\left((H-h)cosθ sinθ+\sqrt{(H-h)^2sin^2θ cos^2θ+(H-h)h cos^2θ}\right)=2cosθ\left((H-h) sinθ+\sqrt{(H-h)^2sin^2θ +(H-h)h}\right)=2cosθ\left((H-h) sinθ+\sqrt{(H-h)(H sin^2θ-hsin^2θ +h)}\right)=2cosθ\left((H-h) sinθ+\sqrt{(H-h)(H sin^2θ+h(1-sin^2θ))}\right)=2cosθ\left((H-h) sinθ+\sqrt{(H-h)(H sin^2θ+h cos^2θ)}\right)$

The interesting fact is that distance of the system doesn't depend on gravity. That means, Azmain would travel same distance in both the Earth and the moon, if there was no wind resistance.
$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$

Post Reply