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NEET PHYSICSEasy

A bullet is fired from a gun at the speed of 280 m s1280 \text{ m s}^{-1} in the direction 3030^{\circ} above the horizontal. The maximum height attained by the bullet is: (g=9.8 m s2,sin30=0.5g=9.8 \text{ m s}^{-2}, \sin 30^{\circ}=0.5)

A

3000 m3000 \text{ m}

B

2800 m2800 \text{ m}

C

2000 m2000 \text{ m}

D

1000 m1000 \text{ m}

Step-by-Step Solution

The maximum height hmh_m reached by a projectile is given by the formula: hm=(v0sinθ0)22gh_m = \frac{(v_0 \sin \theta_0)^2}{2g} where v0v_0 is the initial speed, θ0\theta_0 is the angle of projection, and gg is the acceleration due to gravity .

Given: Initial speed, v0=280 m s1v_0 = 280 \text{ m s}^{-1} Angle of projection, θ0=30\theta_0 = 30^{\circ} Acceleration due to gravity, g=9.8 m s2g = 9.8 \text{ m s}^{-2} sin30=0.5\sin 30^{\circ} = 0.5

Substituting these values into the equation: hm=(280×0.5)22×9.8h_m = \frac{(280 \times 0.5)^2}{2 \times 9.8} hm=(140)219.6h_m = \frac{(140)^2}{19.6} hm=1960019.6h_m = \frac{19600}{19.6} hm=1000 mh_m = 1000 \text{ m}

Thus, the maximum height attained by the bullet is 1000 m1000 \text{ m}.

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