Question

A body is projected with an initial velocity 20 m/s at 60° to the horizontal. the displacement after 2s is​

Answer 1

Explanation:

Horizontal displacement travelled

x = velocity along horizontal × Time

velocity along horizontal at any time is constant (uCos∅)

= (uCos∅) × t

= 20Cos60 × 2

= 20 m

Vertical component of velocity at t = 2 s

V = (uSin∅) - gt

= (20Sin60) - (10 × 2)

= 10√3 - 20

= 10(√3 - 2) m/s

Vertical displacement (i.e height) after 2 s

h = Vt - 0.5gt²

= (10(√3 - 2))2 - (0.5 × -10 × 2²)

= 20(√3 - 2) + 20

= 20√3 - 20

= 14.64 m

Net Displacement

= x² + y²

= (20 m)² + (14.64 m)²

= 614.3 m

Answer 2

Answer:

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