Question

A particle is projected from ground with velocity 40(sqrt2) m//s at 45^@. Find (a) velocity and (b) displacement of the particle after 2 s. (g = 10 m//s^2)

Answer 1

Answer:

(A)  v = usin∅  + at

      v = 40√2 × 1/√2  + 10 × 2

         = 40 + 20 = 60 m/s

(b)  s = usin∅t + 0.5at^2                  

        = 40√2 × 1/√2 × 2 + 0.5 × 10 × 2^2

        = 80 + 20 = 100 m

Answer 2

(a) velocity = 44.72 m/s

(b) displacement of the particle after 2 s

= 100 meter.

Given:

A particle is projected from ground with velocity 40$\sqrt2$ m//s

at 45°,     g = 10 m/s²

To find:

(a) velocity

(b) displacement of the particle after 2 s.

Formula used:

v=u +at

s =ut +$\frac{1}{2}$ at²

where u = initial velocity

          t = time

          a = acceleration

Explanation:

Velocity 40$\sqrt2$ m//s  at 45°        

Velocity in x direction  = 40$\sqrt2$ × cos 45°

                                     = 40 m/s

Velocity in x direction  = 40$\sqrt2$ × sin 45°

                                     = 40 m/s

Velocity vector = $40 \hat{\imath}+40 \hat{\jmath}$

Acceleration vector = $-10 \hat{\jmath}$

                     t = 2 sec.

    $\vec{v}=\vec{u}+\vec{a} \mathrm{t}$

    $\vec{v}=$ ($40 \hat{\imath}+40 \hat{\jmath}$) + ($-10 \hat{\jmath}$) ×2

    $\vec{v}=$ $40 \hat{\imath}+20 \hat{\jmath}$

Magnitude of velocity = $\sqrt{40^{2} + 20^{2} }$ = 44.72 m/s

   $\vec{s}$=$\vec{u}$t +$\frac{1}{2}$ $\vec{a}$t²

   $\vec{s}$ = $80 \hat{\imath}+60 \hat{\jmath}$

Magnitude of displacement = $\sqrt{80^{2} + 60^{2} }$ = 100 meter.

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