The angle which the velocity vector of a projectile, thrown with a velocity $v$ at an angle $θ$ to the horizontal, will make with the horizontal after time $t$ of its being thrown up is

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Answer 1

Horizontal velocity of projectile after time t $= v c o s θ$
Vertical velocity of projectile after time t $= v s i n θ - g t$
Angle at any time t:
$tan ϕ = \frac{v _{v}}{v _{h}} = \frac{v s i n θ - g t}{v c o s θ} ϕ = t a n^{- 1} \frac{v s i n θ - g t}{v c o s θ}$

Answer 2

Horizontal velocity of projectile after time t

= v c o s θ

Vertical velocity of projectile after time t

= v s i n θ - g t

Angle at any time t:

tan ϕ = \frac{v _{v}}{v _{h}} = \frac{v s i n θ - g t}{v c o s θ} ϕ = t a n^{- 1} \frac{v s i n θ - g t}{v c o s θ}

The angle which the velocity vector of a projectile, thrown with a velocity $v$ at an angle $θ$ to the horizontal, will make with the horizontal after time $t$ of its being thrown up is

$θ$

$t a n^{- 1} (\frac{θ}{t})$

$t a n^{- 1} (\frac{v c o s θ}{v s i n θ - g t})$

$t a n^{- 1} (\frac{v s i n θ - g t}{v c o s θ})$

Horizontal velocity of projectile after time t $= v c o s θ$
Vertical velocity of projectile after time t $= v s i n θ - g t$
Angle at any time t:
$tan ϕ = \frac{v _{v}}{v _{h}} = \frac{v s i n θ - g t}{v c o s θ} ϕ = t a n^{- 1} \frac{v s i n θ - g t}{v c o s θ}$

A body is projected at an angle $θ$ with horizontal. Another body is projected with the same velocity at an angle $θ$ with the vertical. The ratio of the time of flights is-

A particle is projected from ground with velocity $v$ at angle $θ$ to horizontal. Average velocity of particle between point of projection and maximum height of projectile is-

A ball is thrown with a velocity $u$ making an angle ' $θ$ ' with the horizontal. Its velocity vector is normal to initial velocity vector $u$ after a time interval of

lf a particle is projected with speed $u$ from ground at an angle $θ$ with horizontal, then radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by :

Two projectiles A and B are thrown with the same speed such that A makes angle $θ$ with the horizontal and B makes angle $θ$ with the vertical, then

Revise with Concepts

Equations of Motion for a Projectile

Maximum Height, Time of Flight and Horizontal Range of a Projectile

Velocity and Direction of Motion at Different Location of Projectile Motion

Trajectory of a Projectile

Equations of Motion for a Projectile - Problem L2

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

θ

tan−1(tθ​)

tan−1(vsinθ−gtvcosθ​)

tan−1(vcosθvsinθ−gt​)

Horizontal velocity of projectile after time t =vcosθ Vertical velocity of projectile after time t=vsinθ−gt Angle at any time t: tanϕ=vh​vv​​=vcosθvsinθ−gt​ϕ=tan−1vcosθvsinθ−gt​

A body is projected at an angle θ with horizontal. Another body is projected with the same velocity at an angle θ with the vertical. The ratio of the time of flights is-

A particle is projected from ground with velocity v at angle θ to horizontal. Average velocity of particle between point of projection and maximum height of projectile is-

A ball is thrown with a velocity u making an angle 'θ' with the horizontal. Its velocity vector is normal to initial velocity vector u after a time interval of

lf a particle is projected with speed u from ground at an angle θ with horizontal, then radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by :

Two projectiles A and B are thrown with the same speed such that A makes angle θ with the horizontal and B makes angle θ with the vertical, then

Revise with Concepts

Equations of Motion for a Projectile

Maximum Height, Time of Flight and Horizontal Range of a Projectile

Velocity and Direction of Motion at Different Location of Projectile Motion

Trajectory of a Projectile

Equations of Motion for a Projectile - Problem L2

The angle which the velocity vector of a projectile, thrown with a velocity $v$ at an angle $θ$ to the horizontal, will make with the horizontal after time $t$ of its being thrown up is

$θ$

$t a n^{- 1} (\frac{θ}{t})$

$t a n^{- 1} (\frac{v c o s θ}{v s i n θ - g t})$

$t a n^{- 1} (\frac{v s i n θ - g t}{v c o s θ})$

Horizontal velocity of projectile after time t $= v c o s θ$
Vertical velocity of projectile after time t $= v s i n θ - g t$
Angle at any time t:
$tan ϕ = \frac{v _{v}}{v _{h}} = \frac{v s i n θ - g t}{v c o s θ} ϕ = t a n^{- 1} \frac{v s i n θ - g t}{v c o s θ}$

A body is projected at an angle $θ$ with horizontal. Another body is projected with the same velocity at an angle $θ$ with the vertical. The ratio of the time of flights is-

A particle is projected from ground with velocity $v$ at angle $θ$ to horizontal. Average velocity of particle between point of projection and maximum height of projectile is-

A ball is thrown with a velocity $u$ making an angle ' $θ$ ' with the horizontal. Its velocity vector is normal to initial velocity vector $u$ after a time interval of

lf a particle is projected with speed $u$ from ground at an angle $θ$ with horizontal, then radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by :

Two projectiles A and B are thrown with the same speed such that A makes angle $θ$ with the horizontal and B makes angle $θ$ with the vertical, then