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xwrathbringerx

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Gee Ming the golfer hits a ball from level ground with an initial speed of 50m/s and an initial angle of elevation of 45 degrees. The ball rebounds off an advertising hoarding 75 m away. Take g = 10m/s^2

a) Show that the ball hits the hoarding after 3/2 * sqrt(2) s at a pt 52.5 m high.

b) Show that the speed v of the ball when it strikes the hoarding is 5 sqrt(58) m/s at an angle of elevation alpha to the horizontal, where alpha = arctan (2/5).

c) Assuming that the ball rebounds off the hoarding at an angle of elevation alpha witha speed of 20% of v, find how far from Gee ming the ball lands.

How exactly do i do c???

By the way, the answer's meant to be 50 m<!-- google_ad_section_end -->
 

Lukybear

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Are you sure the speed is 20% of v?

Cause if answer is 50m, then ball travel 25m after rebounding of board.

Plug that into cartesian and youll get smthing like v=sqrt8

At least for me.
 

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