A 2.3 kg ball is thrown upwards at a speed of 15.0 m/s. What is its initial kinetic energy, how far can it go against gravity before it loses all of its energy, and what is it speed when it is 4.0 m above its starting point?

Question

anonymous · Answer

I'm pretty sure that the answer to the first one is 258.75 J (b/c Ek = o.5mv^2), but I'm not sure how to tackle the other two.

anonymous · Answer

@douglaswinslowcooper

anonymous · Answer

a.  initial KE = (1/2) m v^2


b) final P.E. = m g h = initial K.E. = (1/2) m g h

c). (1/2) m v^2 + m g h = K.E. + P.E. = constant = initial K.E.

set h = 4m

anonymous · Answer

Oh, right. I feel a bit silly now, it appears like I completely forgot about proper formula use just barbecue all the examples beforehand provided the height. Thank you.

anonymous · Answer

I've had that feeling sometimes, too! You are welcome.