Projectile strikes a wall

$\begin{aligned} d_\text{horizontal} &=& \frac{3}{2} \cdot l = 12\sqrt{3}\, \text{m} \\ v_\text{horizontal} &=& \frac{d_\text{horizontal}}{t} = \frac{12\sqrt{3}}{t} \text{m}/\text{s} \end{aligned}$

$\left. \begin{aligned} v_{\text{vertical}} &=& \sin30^\circ \cdot v = \frac{v}{2} \text{m}/\text{s} \\ v_{\text{horizontal}} &=& \cos30^\circ \cdot v = \frac{\sqrt{3}v}{2} \text{m}/\text{s} \end{aligned} \quad \right\} \quad v_{\text{vertical}} = \frac{v_\text{horizontal}}{\sqrt{3}} = \frac{12}{t} \text{m}/\text{s}$

$\begin{aligned} v_\text{vertical} &=& g \cdot t \implies \frac{12}{t} = g \cdot t \implies t^2 = \frac{12}{g} \text{m}^2/\text{s}^2 \\ h &=& \frac{1}{2} gt^2 = \frac{1}{2}g \cdot \frac{12}{g} = \boxed{6}\, \text{m} \end{aligned}$

using the equation of the parabola ( Since the collision is elastic, We can "mirror" the lower part of the parabola to the other side of the wall making it continuous )

y=ax^2+bx+c

y'=2ax+b

origin at the base of tower, ( c=H )

at x=0, y'=0 (because it is thrown horizontally)

at x=12sqrt3 (ground) , y'=-tan(30)

we will get b=0 and a=-1/72

equation of the trajectory --> y=(-x^2)/72 + H

finally at x=12sqrt3 , y=0

H=6m

The velocity at the point just before the ball strikes the ground is given by $V_x$ for its horizontal component of its velocity, and $V_y$ for its vertical component of its velocity.

Hence, by forming a right-angled triangle, we have $tan(30^o) = \frac{V_y}{V_x}$

Note that upon collision with the wall, only the horizontal velocity changes direction, completely (as the collision is elastic)

Let the velocity of the ball just after being thrown be $U_x$ . Hence, $V_x$ = $-U_x$ .

Furthermore, by the kinematics equation $S_x = (U_x)(t)$ , we get $12\sqrt{3} = (-V_x)(t)$ , where $t$ is the time taken for the ball to hit the ground. Hence, $V_x = -\frac{12\sqrt{3}}{t}$

In addition, by the kinematics equation $V_y = U_y + at$ , we have : $V_y = (-g)t$ .

With the expressions for $V_y$ , $U_y$ and that $tan(30^o) = \frac{1}{\sqrt{3}}$ , substitute these expressions into the first equation obtained to get

$\frac{1}{\sqrt{3}} = \frac{gt^2}{12\sqrt{3}}$ , and hence, $12 = gt^2$

With that, to calculate the height of the tower $H$ , where $H = S_y$ , we use $S_y = (U_y)(t) + \frac{1}{2}at^2$ , to get

$H = 0.5gt^2$

Hence, substituting $gt^2 = 12$ , $H = \boxed{6}m$

Energy is not lost anywhere so $\frac{1}{2}mv_0^{2}=mgH$ hence $H=\frac{v_0^{2}}{2g}$

The velocity vector hitting the ground at $30^\circ$ makes a $30^\circ-60^\circ-90^\circ$ triangle with the wall. Therefore for the $60^\circ$ triangle at the wall,

$sin60^\circ=vf_x/vf_y=v_0/vf_y$

where $vf_y$ is the $y$ component of the velocity vector hitting the wall. We can also write $vf_y$ as $vf_y=0 \times t_f + gt_f=gt_f$ where $t_f$ is the time taken to reach the wall. So we have:

$V_0=Vf_ysin60^\circ$

$Vf_y=gt_f$

$t_f=\frac{ 8 \sqrt{3} }{v_0}$ as well, since the horizontal component of $v_0$ remains constant

solving for $V_0$ gives $V_0^{2}=8\sqrt{3}gsin60^\circ$ .Putting $V_0^{2}$ in $H=\frac{V_0^{2}}{2g}$ gives $H=4\sqrt{3}sin60^\circ=6$ .

the relation between H and R is written as H = xtan(thita) (1-x/R) where h=height , R = Range , x=Distance traveled thus H = 4sqrt3 tan30 (1-4sqrt3/8sqrt3) we get H = 2 repeat the same method for upper part of building

if wall would not have been there ball would have taken a symmetric path falling 8sqrt{3}+4sqrt{3) and angle would also remain same ie

tan(theta)=\sqrt { 2gh } /u

and u*t=12\sqrt { 3 }

t=\sqrt { 2h/g }

substituting we get h=6

Note that the wall has no effect on vertical motion, and only serves to reflect the ball back, so our problem is reduced to finding $H$ when there is no wall and the ball instead lands $12\sqrt{3}~m$ horizontally from the tower. If we let the speed the ball was thrown at be $u$ , then because it lands at $30^{\circ}$ to the horizontal, its vertical speed at that time is $\frac{u}{\sqrt{3}}$ .

We can now use SUVAT on the height fallen to deduce that $u^2=58.8H$ $(1)$ (taking $g$ to be $9.8~ms^{-2}$ ) and that $H=4.9t^2$ $(2)$ , where $t$ is the time taken for the ball to fall. We can also deduce using $\text{speed}=\frac{\text{distance}}{\text{time}}$ on the horizontal motion that $u=\frac{12\sqrt{3}}{t}$ $(3)$ . We can then solve equations $(1), (2)$ and $(3)$ to give $H=6~m$ .

horizontal velocity : Vx

vertical velocity : Vy

total time of flight: T

1)Ball is thrown horizontally from the top of tower

(let Vx = u & Vy = gt )

2)Ball collides to the wall elastically.(at a distance 8√3 m away from tower)

since the collision is elastic , ball's horizontal velocity is reversed in direction at the point of collision without any change in its magnitude

(velocity along the wall(Vy) is unaltered since it does not participate in collision of ball to the wall)

(Vx = -u & Vy = gt)

3)ball lands 4√3 m away from the wall

----->net horizontal distance travelled ={8√3 m from tower to wall)}+ {4√3 m from wall to ground} =12√3 m

since horizontal velocity is constant --->horizontal velocity x total time of flight =12√3

---->u xT = 12√3---------(1)

4)Ball lands at an angle of 30° to the ground====>tan30° = Vy/Vx

Vy =gt = g xT =gT & Vx = -u

====>tan 30 = (gT) /(-u) ===>1/√3 = -gT/u ----(2)

we know free fall distance ,h =1/2 gT²
---(since the ball has no vertical velocity component at starting point)

we now know ,uT=12√3 ---(1) & -gT/u = 1/√3 ----(2)

==>(1)x(2) -----> uT x -gT/u =12√3 x 1/√3

===> -gT² = 12

since ,h=1/2 gT² = 1/2 . ( -12) = -6 m

===> free fall distance ,h = -6m

free fall distance = - heigt of tower

so height of tower = 6m

U_{x} = initial horizontal velocity

U_{Y} = initial vertical velocity

V_{x} = final horizontal velocity

V_{Y} = final horizontal velocity

H = height of wall

since it is elastic collision the speed does not change. Only the direction of speed of changes. If you resolve the velocity of ball into horizontal and vertical component, the vertical velocity which is parallel to wall does not change. The horizontal velocity which is perpendicular to wall will simply change its direction (magnitude remains same) after collision. The ball would have hit ground at a distance of 12\sqrt{3} m from wall.

now

tan(30)= V {Y}/V {X} = 1/sqrt{3} ---1

horizontal distance = horizontal velocity * time

12\sqrt{3} = U_{X} \times t ---2

from equation V=U + at (here initial vertical velocity is zero)

V_{y} = gt ---3 (here acceleration of gravity)

from equation V^{2} - U^{2} = 2as (referring to vertical velocity,here a is g, U is zero and s is height of building )

V_{y}^(2) = 2gH --- 4

also U {x} = V {x} ---5 (since no acceleration in horizontal direction)

solve these 5 equations and find H