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Acceleration Free fall Projection

Horizontal downward projection


*Input 3 values in any allowed fields
   
System of units:


Run
Initial incline velocity  (V0 ):
Projection initial angle  (θ0 ):
Horizontal velocity  (Vx ):
Vertical initial velocity  (V0y ):
Projection distance  (S):
Total travel time  (T):
Vertical distance  (H):
Vertical final velocity  (Vt ):
Vertical final velocity  (Vty ):
Projection final angle  (θt ):
Projection path equation:
Input limit:

The downward projection combines two motions, constant speed in the x direction and free fall in the downward y direction. The motion equations are:

In the x direction:
S=V_(x ) T
In the y direction:
H=V_y0 T+1/2 gT^2
The path of the downward projection describes the right half of the parabola and is:
(the equation is derived from the general equation of motion of a projection θ0 = 0)
y=g/(2〖V_0〗^2 ) x^2

The signs of the variables should follow the rules

Positive values:
S
T
Vx
V0
Vt
Negative values:
θ0
θt
V0y
Vty
H
g
img2Show

Example of detailed solution

An object is launched downward at an angle of -35° degree and initial velocity of 12 m/s. Find all the parameters at the point the object is at a velocity of 25 m/s.
From the question we have the values of:

V_0=12 m/s  θ_0=-30°  V_t=25  m/s

The horizontal velocity is equal to:

V_x=V_0  cos⁡〖θ_0 〗=12 cos⁡〖30=10.39 m/s〗

The downward initial velocity is calculated from:

V_0y=V_0  sin⁡〖θ_0 〗=12 sin⁡〖30=6 m/s〗

The downward final velocity is calculated from:

V_ty=√(〖V_t〗^2-〖V_x〗^2 )=√(25^2-〖10.39〗^2 )=22.74  m/s

Next we can find the value of the time T:

T=√((V_ty-V_0y)/(-g))=√((-22.74+6)/(-9.8))=1.31 s

Now we can find the horizontal distance made by the object:

S=V_x T=10.39∙1.31=13.61 m

The total vertical drop height made by the object:

H=V_0y T+1/2 gT^2=-6∙1.31-1/2 9.8∙〖1.31〗^2=-16.27 m

Finaly the angle at the end point is:

θ_t=〖tan〗^(-1)⁡〖V_ty/V_x 〗=tan^(-1)⁡〖(-22.74)/10.39〗=-65.44°

Horizontal downward projection - Detailed solutions of various inputs in all cases Vx was found
V0   H   S
V0   Vy   S
V0   θy   H
V0   θy   T
V0   θy   S
Vy0   θy   S
Vy   θ0   S
θ0   θy   H
θ0   θy   T
θ0   θy   S
θ0   H   S
θy   H   S
Vy  θ0  -> Smax
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