Posted by Ted on .
A stone is thrown straight up. To reach a height of 6.49 m above its release point, what must its initial speed be in m/s?

Physics 
bobpursley,
initial KE=final PE
1/2 mv^2=mg(6.49)
solve for v
Posted by Ted on .
A stone is thrown straight up. To reach a height of 6.49 m above its release point, what must its initial speed be in m/s?
initial KE=final PE
1/2 mv^2=mg(6.49)
solve for v