A cannon at the ground fires a shell at speed of 50 km/hr. If the angle of projection is 30° , the range of the shell (take g=10 text m/s2 ) is, in km.

Question

A cannon at the ground fires a shell at speed of 50 km/hr. If the angle of projection is 30° , the range of the shell (take g=10 text m/s2 ) is, in km. (A)  2.5√3  (B)  62.5√3  (C)  125√3  (D)  9.65√3× 10-3

Tardigrade Admin · Accepted Answer

: Range of projectile =R=(u2 sin 2θ /g) ∴ R=( (50× 1000/60× 60) )2× ( sin 60° /10) Or R=(125× 125× √3/9× 9× 2× 10)m Or R=9.65√3m Or R=9.65√3× 10-3km

Q. A cannon at the ground fires a shell at speed of 50 km/hr. If the angle of projection is $30^{\circ},$ the range of the shell (take $g = 10 m / s^{2}$ ) is, in km.

$2.5 3$

$62.5 3$

$1253$

$9.65 3 \times 10^{- 3}$

: Range of projectile $= R = \frac{u ^{2} s i n 2 θ}{g}$ $∴$ $R = (\frac{50 \times 1000}{60 \times 60})^{2} \times \frac{s i n 60 ^{\circ}}{10}$ Or $R = \frac{125 \times 125 \times 3}{9 \times 9 \times 2 \times 10} m$ Or $R = 9.65 3 m$ Or $R = 9.65 3 \times 10^{- 3} km$

Q. A cannon at the ground fires a shell at speed of 50 km/hr. If the angle of projection is 30∘, the range of the shell (take g=10m/s2 ) is, in km.

2.5​3

62.5​3

125​3

9.65​3×10−3