How do I determine the components of a cinematic jump, for vertical and horizontal velocity? I have been tasked with determining the feasibility of The Rock's jump in the movie 'Skyscraper' I am using projectile motion equations to determine it, but have gotten stuck whilst calculating my horizontal and vertical velocity. However, the values I have been getting have not been close to satisfying a vector sum for the total velocity.
My calculations are listed as below:
Distance Run= 26 metres
Time Taken= 3.03 seconds
Horizontal Velocity = distance/time = 36/3.03 = 8.58m/s
I calculated the angle of the jump to be 70 degrees, from a snippet of the movie, and the maximum height of the jump to occur after 0.73 seconds

Velocity= vsinangle + gravity* time
0=v*sin70 -9.8*0.73
7.15=v*sin*70
v= 7.61m/s
The sum of the vectors is 16.19m/s, which does not satisfy a Pythagorean triangle. 
I am wondering where I have gone wrong, and what I have to do to solve for the initial velocity. 
Thankyou.
 A: Of course the three numbers don't describe the sides of a right triangle. You are incorrectly assuming that $v_0=v_{0,x}+v_{0,y}$
and then also expecting the Pythagorean theorem to work out with the then incorrect value of $v_0$. This makes sense because, in general,
$$v_{0,x}+v_{0,y}\neq\sqrt{v_{0,x}^2+v_{0,y}^2}$$
The sum of your vectors should be a vector sum
$$\mathbf{v}_0=v_{0,x}\hat i +v_{0,y}\hat j$$
Whose magnitude is then given by
$$v_0=\sqrt{v_{0,x}^2+v_{0,y}^2}$$
And whose angle is given by
$$\tan\theta=\frac{v_{0,y}}{v_{0,x}}$$
These equations should then be self consistent between your measured and calculated values. Then you will be able to smell what the math is cooking.
A: The initial vertical velocity $v_{y0}$ can be found by using $$0 = v = v_{y0} - gt.$$ The initial horizontal velocity is going to be the sum of the velocity due to the jump and the velocity due to the run. You correctly solved for the horizontal velocity due to the run: $v'_{x0} = 8.58 \, \textrm{m/s}$. The horizontal velocity due to the jump $v''_{x0}$ is given by $$\tan\theta = \frac{v_{y0}}{v''_{x0}}.$$
