derive the velocity of the body at any time 't' in a projectile motion

Open in App

Solution

Ans: As we know that in projectile motion the components of velocity and acceleration are calculated in both the horizontal and vertical direction. so if the the object is project with 'u' initial velocity and making angle ' $θ$ ' with horizontal, and attains
velocity 'v' after time 't', and the direction of motion is along the line making angle ' $α$ ' with horizontal.
acceleration of the body is determined by acc. due to gravity 'g' which acts along verticle direction, and no acc. is there along horizontal direction.
(1) Horizontal component of the velocity:
$A c c o r d i n g t o t h e f i r s t e q u a t i o n o f m o t i o n : v = u + a t v \cos (α) = u \cos (θ) + 0 \times t v c o s (α) = u c o s (θ) . . . . . . (1)$
(2) Verticle component of the velocity:
$v \sin (α) = u \sin (θ) - g t . . . . (2) [a c c . i s o p p o s i t e t o t h e d i r . o f m o t i o n] n o w p e r f o r m i n g f o l l o w i n g f u n c t i o n : . . (1)^{2} + . . . (2)^{2} [v^{2} \cos {(α)}^{2} + v^{2} \sin {(α)}^{2}] = u \cos {(θ)}^{2} + [u s i n (θ) - g t]^{2} v^{2} [c o s {(α)}^{2} + s i n {(α)}^{2}] = u^{2} c o s {(θ)}^{2} + u^{2} \sin {(θ)}^{2} + g^{2} t^{2} - 2 g t u \sin (θ) v^{2} = u^{2} [c o s {(θ)}^{2} + u^{2} s i n {(θ)}^{2}] + g^{2} t^{2} - 2 g t u s i n (θ) v = \sqrt{u^{2} - 2 g t u s i n (θ) + g^{2} t^{2}} : A n s$

Suggest Corrections

Similar questions

Q. Derive formula for the time period of projectile motion.

Q. In a projectile motion let

v_{x}

and

v_{y}

be the horizontal and vertical components of velocity at any time

t

and

x

and

y

be the displacements along horizontal and vertical from the point of projection at any time

t

. Then

Statement1: In absence of air, projectile motion is always a 2D motion.
Statement2: In absence of any external force in 3D space, at any time during its flight Projectile's velocity has only two components.

Q. A projectile is fired with a velocity

u

at angle

θ

with the ground surface. During the motion at any time it is making an angle

α

with the ground surface. The speed of particle at this time will be

Q. Derive the equations of projectile motion.

Join BYJU'S Learning Program