Projectile Motion - What's the Range(R)?

Take g = 9.81 m/s², and take all downward quantities as positive in vertical movement.

Vertical displacement : Horizontal displacement = 3 : 4
Hence, Vertical displacement = (3/4) * Horizontal displacement = 0.75R


Consider the vertical motion (uniform acceleration motion) :
Initial velocity, u = -10 sin40° m/s
Displacement, s = 3R/4
Acceleration, a = 9.81 m/s²

s = ut + (1/2)at²
0.75R = (-10 sin40°)t + (1/2)(9.81)t² …… [1]

Consider the horizontal motion (uniform velocity motion) :
Displacement = Horizontal velocity × time
R = (10 cos40°)t …… [2]

Substitute [2] into [1] :
(0.75)(10 cos40°)t = (-10 sin40°)t + (1/2)(9.81)t²
4.905t² = (7.5 cos40°)t + (10 sin40°)t

For t ≠ 0 :
4.905t = (7.5 cos40°) + (10 sin40°)
t = [(7.5 cos40°) + (10 sin40°)]/4.905
t = 2.48 s

Substitute t = 2.48 s into [2] :
R = (10 cos40°) * 2.48 m
The range, R = 19 m

R = Ux T = 10*cos(40)*T; we need T, the flight time.

So we have y = 0 = h + UyT - 4.9T^2; where h = 3/4 R is the height above impact at B. So 4.9T^2 - UyT - 3/4 R = 0 = 4.9T^2 - UyT - .75 Ux T then

4.9T - Uy - .75 Ux = 0
4.9T = Uy + .75 X
T = (Uy + .75 Ux)/4.9 = (10*sin(radians(40)) + .75*10*cos(radians(40)))/4.9 = 2.48 seconds flight time.

Then R = Ux T = 10*cos(radians(40)))*2.48 = 1.8998E+01 ANS.