Alan

Assume he gets heads h times; then he must get tails 20-h times. So

5*h - 3*(20-h) = 4

5h - 60 +3h = 4

8h - 60 = 4

8h = 64

h = 8

He threw heads 8 times.

The useful formulae here are the constant acceleration kinematics formulae, namely:

1.  v = u + at

2.  v² = u² + 2as

3.  s = ut + (1/2)at²

u is initial velocity, v is final velocity, a is acceleration, t is time, s is distance. 

So, we can use equation 2. to find the final velocity of the ball thrown vertically down (assuming no resistance due to air drag):

v² = 13.1² + 2*9.8*18.8

$${\mathtt{v}} = {\sqrt{{{\mathtt{13.1}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{18.8}}}} \Rightarrow {\mathtt{v}} = {\mathtt{23.239\: \!836\: \!488\: \!237\: \!175\: \!2}}$$

So, velocity of first ball hitting ground is approximately 23.24m/s.

For the second ball the velocity as it hits the ground will be the same as it will lose speed as it rises and then regain the same speed by the time it gets back to the balcony, from where it will behave like the first ball to the ground!

To find the times use equation 3. for the first ball. and, for the second ball, equation 1. (setting v = 0) for the upward part of the flight and 3. for the downward part.

Here's the reasoning:

θ is the angle whose cosine is 3*√(6)/14 so

θ = cos^-1(3*√(6)/14)

$${\mathtt{theta}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}}}}{{\mathtt{14}}}}\right)} \Rightarrow {\mathtt{theta}} = {\mathtt{58.339\: \!117\: \!225\: \!405^{\circ}}}$$

180° is pi radians, so in radians we have

$${\mathtt{theta}} = {\frac{{\mathtt{58.339\: \!117\: \!225\: \!405}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}} \Rightarrow {\mathtt{theta}} = {\mathtt{1.018\: \!209\: \!678\: \!290\: \!256\: \!2}}$$

or θ ≈ 1.02 radians

This seems to refer to the well-known right-angled triangle with sides BC = 3, AC = 4, AB = 5, in which case cos(CAB) = 4/5 (=0.8)

1. Find the distance from A to the spot on the ground immediately below the balloon (i.e. point (0, 0, 0)) - I'll call this spot E.  Use Pythagoras theorem to do this.  AE = √(5² + 9²).

2.  The force along EA is F_EA = F_ADcos(A), where A = tan^-1(20/AE)

3.  Resolve F_EA along the x and y directions.  F_EAx = F_EA*5/AE,  F_EAy = -F_EA*9/AE

4.  Replace F_EA in step 3 by F_ADcos(A).

5.  Do the same for the forces F_BD and F_CD.

6.  Resolve F_AD, F_BD, and F_CD along the z-direction.

7.  Balance the forces in the x-direction.  Balance the forces in the y-direction.  Balance the forces in the z-direction (don't forget the vertical force on the balloon).  This should give you three equations for the three unknown forces F_AD, F_BD, and F_CD.

Let F_AB be the force in cable AB and F_BC be the force in cable BC.  Let θ_A be the angle between cable AB and the horizontal and θ_C be the angle between cable BC and the horizontal.

Horizontal force balance

F_ABcos(θ_A) = F_BCcos(θ_C)          ...(1)

Vertical force balance

F_ABsin(θ_A) = mg + F_BCsin(θ_C)  ...(2)

m = 2000kg; g = 9.81m/s².

We also have that θ_A = tan^-1(5.6/2.2) and θ_C = tan^-1(2.4/5.6)

Replace F_BC in (2) by using (1)

F_ABsin(θ_A) = mg + F_ABcos(θ_A)sin(θ_C)/cos(θ_C)

Rearrange to get F_AB

F_AB = mg/(sin(θ_A) - cos(θ_A)sin(θ_C)/cos(θ_C))

$${\mathtt{FAB}} = {\frac{{\mathtt{2\,000}}{\mathtt{\,\times\,}}{\mathtt{9.81}}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.6}}}{{\mathtt{2.2}}}}\right)}\right)}{\mathtt{\,-\,}}{\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.6}}}{{\mathtt{2.2}}}}\right)}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.4}}}{{\mathtt{5.6}}}}\right)}\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.4}}}{{\mathtt{5.6}}}}\right)}\right)}}}\right)}} \Rightarrow {\mathtt{FAB}} = {\mathtt{25\,347.418\: \!086\: \!936\: \!201\: \!807\: \!8}}$$

FAB ≈ 25.347 kN

Substitute this back into (1) to get F_BC

F_BC = F_ABcos(θ_A)/cos(θ_C)

$${\mathtt{FBC}} = {\frac{{\mathtt{25\,347.418}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.6}}}{{\mathtt{2.2}}}}\right)}\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.4}}}{{\mathtt{5.6}}}}\right)}\right)}}} \Rightarrow {\mathtt{FBC}} = {\mathtt{10\,083.657\: \!337\: \!858\: \!801\: \!218\: \!7}}$$

FBC ≈ 10.083 kN

The ∑ sign usually means sum over a number of terms.  However, you only have one term, namely 10^2.

Have you used the right symbol here?

Slope intercept form is y = mx + c where m is slope and c is y-intercept.

Rewrite your equation as: y = -5x + 5√3 + 2√3 = -5x +7√3

so the slope is -5 and the y-intercept is 7√3

$$$$v = \frac{\pi}{3}r^2h$$\\\\
Multiply both sides by $\frac{3}{\pi h}$ to get $$v\frac{3}{\pi h}=r^2$$\\\\
Take the square root of both sides
$$ r=\sqrt{v\frac{3}{\pi h}}$$