Question

A particle is projected with a velocity of 50 m/s at 37^@ with horizontal. Find velocity, displacement and co-ordinates of the particle (w.r.t. the starting point) after 2 s. Given, g=10m//s^2, sin 37^@ = 0.6 and cos 37^@ = 0.8.

Answer 1

Answer:

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Answer 2

The coordinates of the projectile after two seconds are (10,80)

Horizontal velocity=50m/s

Vertical velocity =30m/s

Displacement of the projectile is 10m

Given the velocity of the projectile is 50 m/s

Angle of projectile is 37°

g=10 m/S2

Coordinates of the projectile after 2 sec x=utsina-0.5*g*t^2

=50*0.6 - 0.5*10*2*2

=10m

Y=utcosa

=50*2*0.8

=80m

(x,y)=(10,80)

Horizontal velocity remains constant where as vertical velocity changes -9.8 for every sec

Displacement of the projectile is the range of the projectile i.e x= 10 m