Figure shows a projectile thrown with speed u=20 m / s at an angle 30° with horizontal from the top of a building 40 m high. Then the horizontal range of projectile is

Question

Figure shows a projectile thrown with speed u=20 m / s at an angle 30° with horizontal from the top of a building 40 m high. Then the horizontal range of projectile is (A) 20 √3 m (B) 40 √3 m (C) 40 m (D) 20 m

Tardigrade Admin · Accepted Answer

Sy=uy T+(1/2) gy T2 -40=4 sin 30 T-(1/2) g T2 -40=20 × (1/2) T-5 T2 -8=2 T-T2 T2-2 T-8=0 T2-4 T+2 T-8=0 T=-2,4 R=u cos θ T=20 × (√3/2) × 4 R=40 √3 m

Q. Figure shows a projectile thrown with speed $u = 20 m / s$ at an angle $3 0^{\circ}$ with horizontal from the top of a building $40 m$ high. Then the horizontal range of projectile is

$203 m$

$403 m$

$40 m$

$20 m$

$S_{y} = u_{y} T + \frac{1}{2} g_{y} T^{2}$
$- 40 = 4 sin 30 T - \frac{1}{2} g T^{2}$
$- 40 = 20 \times \frac{1}{2} T - 5 T^{2}$
$- 8 = 2 T - T^{2}$
$T^{2} - 2 T - 8 = 0$
$T^{2} - 4 T + 2 T - 8 = 0$
$T = - 2, 4$
$R = u cos θT = 20 \times \frac{3}{2} \times 4$
$R = 403 m$

Q. Figure shows a projectile thrown with speed u=20m/s at an angle 30∘ with horizontal from the top of a building 40m high. Then the horizontal range of projectile is

203​m

403​m

40m

20m