One breezy day the wind gives rise to a horizontal component of acceleration equ

Question

One breezy day the wind gives rise to a horizontal component of acceleration equal to 3.7 m/s2. This is in addition to the usual acceleration due to gravity.

If a soccer player on a level field kicks the ball with an initial velocity that is straight up and has a magnitude 8.4 m/s, what is the horizontal distance the ball will travel before returning to the ground? Assume the ball begins its upward motion from the ground.

5.43 m

The player now wishes to kick the ball so that it will land in the same location from which it is now kicked (back at the player's foot). At what angle with respect to the ground should the player kick the ball?

Thank you in advance for the help!

Explanation / Answer

HE should kick in the opposite direction of the wind

For zero displacement, s=0

s=ut+ 0.5at^2

0= ut-1.85t^2

since t cannot be zero, u= 1.85t

He kicks at the speed of 8.4 m/s

u= 8.4 cos theta

v= vertical velocity= 8.4 sin theta

s= 8.4t- 4.9t^2

even the vertical displacement is 0, hence

t= 1.71s

u= 1.85 x 1.71= 3.17m/s

cos theta= 3.17/8.4

theta=67.82 deg

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