Question

The angle which the velocity vector of a projectile thrown with a velocity v at an angle theta to the horizontal will make with yours until after time t of its being thrown up

Answer 1

consider the motion in Y-direction

$V_{oy}$ = v SInθ

a = acceleration = - g where g = 9.8

t = time interval

$V_{fy}$ = final velocity at time "t"

Using the equation

$V_{fy}$ = $V_{oy}$ + a t

$V_{fy}$ = v SInθ - gt

Consider the motion along X-direction

$V_{ox}$ = v Cosθ

a = acceleration = 0

t = time interval

$V_{fx}$ = final velocity at time "t"

Using the equation

$V_{fx}$ = $V_{ox}$ + a t

$V_{fx}$ = v Cosθ

Φ = angle after time "t"

angle is given as

Φ = tan⁻¹ ($\frac{v_{oy}}{v_{ox}}$)

Φ = tan⁻¹ ($\frac{(v Sin\theta - gt)}{v Cos\theta }$)

Answer 2

consider the motion in Y-direction

V_{oy}Voy​ = v SInθ

a = acceleration = - g where g = 9.8

t = time interval

V_{fy}Vfy​ = final velocity at time "t"

Using the equation

V_{fy}Vfy​ = V_{oy}Voy​ + a t

V_{fy}Vfy​ = v SInθ - gt


Consider the motion along X-direction

V_{ox}Vox​ = v Cosθ

a = acceleration = 0

t = time interval

V_{fx}Vfx​ = final velocity at time "t"

Using the equation

V_{fx}Vfx​ = V_{ox}Vox​ + a t

V_{fx}Vfx​ = v Cosθ

Φ = angle after time "t"

angle is given as

Φ = tan⁻¹ (\frac{v_{oy}}{v_{ox}}vox​voy​​ )

Φ = tan⁻¹ (\frac{(v Sin\theta - gt)}{v Cos\theta }vCosθ(vSinθ−gt)​ )