Question #102527

A pebble is shot vertically at 28m/s ignoring air resistance. What height will the pebble reach?


1
Expert's answer
2020-02-10T09:08:02-0500

From the conservation of energy:


0.5mv2=mgh0.5mv^2=mgh

h=v22g=2822(9.8)=40 mh=\frac{v^2}{2g}=\frac{28^2}{2(9.8)}=40\ m


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