A projectile is thrown with speed 40 ms -1 at angle θ from horizontal. It is found that projectile is at same height at 1 s and 3 s. What is the angle of projection?

Question

A projectile is thrown with speed 40 ms -1 at angle θ from horizontal. It is found that projectile is at same height at 1 s and 3 s. What is the angle of projection? (A)  tan -1((1/√2)) (B)  tan -1((1/√3)) (C)  tan -1(√3) (D)  tan -1(√2)

Tardigrade Admin · Accepted Answer

tan θ=(vy/vx) Also, t1+t2=(2 u sin θ/g) 4=(2 × 40 × sin θ/10) sin θ=(1/2) ⇒ θ=30° So, tan θ= tan 30° ⇒ (1/√3) θ= tan -1((1/√3))

Q. A projectile is thrown with speed $40 m s^{- 1}$ at angle $θ$ from horizontal. It is found that projectile is at same height at $1 s$ and $3 s$ . What is the angle of projection?

$tan^{- 1} (\frac{1}{2})$

$tan^{- 1} (\frac{1}{3})$

$tan^{- 1} (3)$

$tan^{- 1} (2)$

$tan θ = \frac{v _{y}}{v _{x}}$
Also, $t_{1} + t_{2} = \frac{2 u s i n θ}{g}$
$4 = \frac{2 \times 40 \times s i n θ}{10}$
$sin θ = \frac{1}{2}$
$\Rightarrow θ = 3 0^{\circ}$
So, $tan θ = tan 3 0^{\circ}$
$\Rightarrow \frac{1}{3}$
$θ = tan^{- 1} (\frac{1}{3})$

Q. A projectile is thrown with speed 40ms−1 at angle θ from horizontal. It is found that projectile is at same height at 1s and 3s. What is the angle of projection?

tan−1(2​1​)

tan−1(3​1​)

tan−1(3​)

tan−1(2​)