Question

a particle is projected at an angle 37 degree from ground with velocity 36 km per hour. the velocity after 1 second is

Answer 1

The velocity will be 34 km/h

Answer 2

Horizontal displacement of the projectile after 't' seconds [t=1] ;

x = (u cos θ) t = u cos 37° = 4u/5 = 28.8 [as u = 36 ]

Vertical displacement of the projectile after 't' seconds [t=1] ;

y = (u sin θ) t - (gt²/2) = (3u/5) - (g/2) = 21.6 - 4.9 = 16.7 [as u = 36 and g = 9.8]

Thus velocity after 1 second = √[ x² + y² ] = √[ 28.8² + 16.7² ] = √[1108.33]

= 33.29

Thus the velocity of the particle after 1 second will be 33.29 km/hr