A projectile is launched at an angle ' α ' with the horizontal with a velocity 20 ms -1. After 10 s, its inclination with horizontal is ' β '. The value of tan β will be: (g=10 ms -2)

Question

A projectile is launched at an angle ' α ' with the horizontal with a velocity 20 ms -1. After 10 s, its inclination with horizontal is ' β '. The value of tan β will be: (g=10 ms -2) (A)  tan α+5 sec α (B)  tan α-5 sec α (C) 2 tan α-5 sec α (D) 2 tan α+5 sec α

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d76bdfe18cf2fa36a505aada012c3c45-.png" /> vx=ux=20 cos α vy=20 sin α-10 × 10 tan β=(vy/vx)=(20 sin α-100/20 cos α) = tan α-5 sec α

Q. A projectile is launched at an angle ' $α$ ' with the horizontal with a velocity $20 m s^{- 1}$ . After $10 s$ , its inclination with horizontal is ' $β$ '. The value of $tan β$ will be : $(g = 10 m s^{- 2})$

$tan α + 5 sec α$

$tan α - 5 sec α$

$2 tan α - 5 sec α$

$2 tan α + 5 sec α$

$v_{x} = u_{x} = 20 cos α$
$v_{y} = 20 sin α - 10 \times 10$
$tan β = \frac{v _{y}}{v _{x}} = \frac{20 s i n α - 100}{20 c o s α}$
$= tan α - 5 sec α$

Q. A projectile is launched at an angle ' α ' with the horizontal with a velocity 20ms−1. After 10s, its inclination with horizontal is ' β '. The value of tanβ will be : (g=10ms−2)

tanα+5secα

tanα−5secα

2tanα−5secα

2tanα+5secα