Question

michael kicks a ball at an angle if 36* horizontal. its initial velocity is 46 m/s. Find the maximum height it can reach, total time, and horizontal displacement for this motion

Answer 1

(a) At its maximum height, the ball's vertical velocity is 0. Recall that

${v_y}^2-{v_{0y}}^2=2a_y\Delta y$

Then at the maximum height $\Delta y=y_{\mathrm{max}}$, we have

$-\left(\left(46\,\dfrac{\mathrm m}{\mathrm s}\right)\sin36^\circ\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}$

$\implies y_{\mathrm{max}}=37\,\mathrm m$

(b) The time the ball spends in the air is twice the time it takes for the ball to reach its maximum height. The ball's vertical velocity is

$v_y=v_{0y}+a_yt$

and at its maximum height, $v_y=0$ so that

$0=\left(46\,\dfrac{\mathrm m}{\mathrm s}\right)\sin36^\circ+\left(-9.8\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$\implies t=2.8\,\mathrm s$

which would mean the ball spends a total of about 5.6 seconds in the air.

(c) The ball's horizontal position in the air is given by

$x=v_{0x}t$

so that after 5.6 seconds, it will have traversed a displacement of

$x=\left(46\,\dfrac{\mathrm m}{\mathrm s}\right)\cos36^\circ(5.6\,\mathrm s)$

$\implies x=180\,\mathrm m$