How to derive this expression step by step?

Question

The equation for horizontal range of a projectile launched from the given height y0 with initial velocity v0 at angle θ is: x=vcosθ/g(vsinθ+(√v²sin²θ+2gy0)

Answer 1

(I updated a few times to just add more description)

 

the distance of the horizontal range will be the horizontal velocity x total time until it hits the ground. 

 

at a height Y0 the projectile is launched at an angle theta.

 

at t = 0, 

1)  VX₀  = V₀*COS(THETA)    <-- HORIZ VELOCITY STAYS CONSTANT

2)  VY₀  = V₀*SIN(THETA)     <-- VERT VELOCITY DECREASES (ACCEL DOWNWARD GRAVITY G)

 

1ST CALC THE TIME AT WHICH IT REACHES PEAK HEIGHT, WHERE VERTICAL VEL = 0

 

Vy (AT PEAK) = 0 = VY₀ - G*T1    (T1 is the delta time to reach peak height)

SO  T1  = -VY₀/-G  

      T1  =  VY₀/G   or using eq. 2 for VY₀

 

3)  T1  = V₀*SIN(THETA)/G     

 

NOW CALC THE ADDED HEIGHT (DY) ABOVE Y0 WHERE IT REACHED IT'S PEAK HT. 

 

    DY = VY₀*T1 - G*T1²/2        (KINEMATICS EQN)

 

SUBSTITUTE EQ. 2,3 FOR VY₀ AND T1 RESP. 

 

    DY = V₀*SIN(THETA)*V₀*SIN(THETA)/G  -  (G/2)*(V0²*SIN²(THETA)/G²)

 

SIMPLIFYING:

 

   DY = V₀²*SIN²(THETA)/G  -  (1/2)*V₀²*SIN²(THETA)/G

 

4) DY = (1/2)*V₀²*SIN²(THETA)/G

 

IT'S TOTAL HEIGHT ABOVE THE GROUND AT THE PEAK IS Y0 + DY

 

WE NOW NEED TO COMPUTE THE DELTA TIME T2  FROM PEAK HT UNTIL IT HITS THE GROUND OR ZERO HT. 

 

Y0 + DY -  G*(T2)²/2 = 0    OR  Y0 + DY = G*(T2)²/2

 

5)  T2² = 2*(Y0 + DY)/G

 

WE NOW SUBSTITUTE DY FROM EQ.4  IN EQ 5. 

 

T2² = 2*(Y0 +  (1/2)*V₀²*SIN²(THETA)/G) / G

 

SIMPLIFYING:

 

T2² = 2*(Y0*G + (1/2)*V₀²*SIN²(THETA)) /G / G

 

T2² = 2*(Y0*G + (1/2)*V₀²*SIN²(THETA)) /G²

 

TAKING THE SQ ROOT TO GET T2:

 

6.  T2 = SQRT(2*Y0*G + V₀²*SIN²(THETA)) /G

 

 

SO NOW, WE HAVE THE TOTAL TIME = T1 + T2

THE HORIZ DISTANCE IS THE HORIZ VELOCITY VX * TOTAL TIME (T1 + T2)

 

DX (HORIZ RANGE) = VX*(T1 + T2)

 

SO WE NEED  VX, T1, AND T2  FROM EQUATIONS 1, 3, 6 

 

 

DX (HORIZ RANGE) = V0*COS(THETA) * V0*SIN(THETA)/G  + SQRT(2*Y0*G + V₀²*SIN²(THETA)) /G

 

SIMPLIFYING, PULLING THE DIVISOR G OUT FRONT WE FINALLY HAVE:

 

DX (HORIZ RANGE) = (V0*COS(THETA)/G) * V0*SIN(THETA) + SQRT(2*Y0*G + V₀²*SIN²(THETA))

 

I BELIEVE THIS MATCHES YOUR EQN. 

 

 

IN REVIEW, WE HAD TO SOLVE FOR DELTA TIMES T1 AND T2 AS THE TOTAL TIME. 

SINCE HORIZ VELOCITY IS CONSTANT, THIS VELOCITY (VX) TIMES THE TOTAL TIME IS THE HORIZONTAL RANGE. 

T1 WAS SOLVED BY FINDING WHEN VERT VELOCITY WENT TO ZERO DUE TO GRAVITY. 

WE THEN FOUND THE DELTA HT ABOVE Y0 BY DY = GT²/2 (T = T1) , FOR A TOTAL HT OF Y0 + DY.

WE THEN FOUND T2, OR THE TIME TO FALL FROM Y0 + DY.  

USING THE SAME KINEMATIC EQN. Y0+DY = GT²/2  (T = T2)    T1+T2 is THEN THE TOTAL TIME OF FLIGHT.