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FinnZ [79.3K]
2 years ago
12

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
Physics
1 answer:
MAVERICK [17]2 years ago
3 0

(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

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