A batter hits a base ball so that it leaves the bat with an initial speed $37\; m/s$ at an angle $53^{\circ}$. Find the position of the ball and the magnitude and direction of its velocity after $2$ seconds. Treat the baseball as a projectile.

Given,

$u = 37\;m/s$

$v = 53^{\circ}$

$t = 2\;sec$

$s = u_y*t - \frac{1}{2}\;g\;t^2$

$s = u\;Sin(\theta) * t - \frac{1}{2}\;g\;t^2$

$s = 37\;Sin (57) * 2 - \frac{1}{2}10 * 2^2$

⇒ $s = 39.5\;m$

And, Horizontal displacement = $u_y * t$

                                      $= u\; Cos\theta * 2 = 37 * Cos(53) * 2 = 44.52\;m$

∴ So, After 2 Sec the baseball will be lying $39.5\;m$ above its point of projection and $44.52\;m$ ahead of its point of projection.

Now, Let the vertical component of velocity will become $v_y$ after time 2 sec

$v_y = u_y -g*t = u\;Sin\theta - g\;t$

$v_y = 9.95 \;m/s$

$v_x = u_x = u\;Cos\theta = 22.27\;m/s$

Magnitude of velocity after $2\;s$ is,

$V = \sqrt{v_x^2 + v_y^2} = \sqrt{22.27^2 + 9.95^2} = 24.4\;m/s$

$\theta = tan^{-1}(\frac{v_y}{v_x}) = 24.06^{\circ}$ with the horizontal.

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