Question

A particle is projected from ground with velocity 50 m//s at 37^@ from horizontal. Find velocity and displacement after 2 s. sin 37^@ = 3/5.

Answer 1

Given:-

Initial velocity u = 50m/sec.

Angular Projection θ = 37°.

Time t = 2sec.

To Find:-

Determine final velocity and displacement.

Solution:-

Then Initial velocity in x and y direction

$u_{x}$ = 50cos37° = 50x(4/5) = $50x\frac{4}{5}$ = 40m/sec.

$u_{y}$ = 50sin37° = 50x(3/5) = $50x\frac{3}{5}$ = 30m/sec.

Initial Velocity Vector $\vec{U}$ = $(40\hat{i} + 30\hat{j})$m/sec

Acceleration Vector  $\vec{a}$ = (-10$\hat{j}$)m/sec².

Final velocity

$\vec{V}$ = $\vec{U}$ + $\vec{a}$t

$\vec{V}$ = $(40\hat{i} + 30\hat{j})$ + (-10$\hat{j}$)(2).

Final velocity V =  $(40\hat{i} + 10\hat{j})$

V = $\sqrt{40^{2} +10^{2} } = \sqrt{1700} = 41.23m/sec$

Final Displacement Vector.

 $\vec{S}$ = $\vec{U}$t + 1/2$\vec{a}$t²

 $\vec{S}$ = $\vec{U}$t + $\frac{1}{2}at^{2}$

 $\vec{S}$ =$(40\hat{i} + 30\hat{j})(2)$ + 1/2(-10$\hat{j}$)(4)

 $\vec{S}$ = $(40\hat{i} + 30\hat{j})(2)$ + $\frac{1}{2}(-10j)(4)$

 $\vec{S}$ = ( 80$\hat{i}$ +40$\hat{j}$)

 Then final Displacement

  $S_{net}$ = $\sqrt{80^{2}+ 40^{2} } = \sqrt{8000} = 89.44m.$

 Final Displacement will be = 89.44m