Aprojectile is launched straight up at 141 m/s . how fast is it moving at the top of its trajectory? suppose it is launched upward at 45 degree above the horizontal plane. how fast is it moving at the top of its curved trajectory?

The velocity of projectile has 2 components, horizontal component vcosθ and vertical component vsinθ, where v is the velocity of projection and θ is the angle between +ve X-axis and projectile motion.

In case 1, θ = 90⁰

    So horizontal component is vcos90 = 0

         Vertical component at maximum height = 0

So velocity at maximum height = 0 m/s

In case 2, θ = 45⁰

    So horizontal component is 141cos45 = 100m/s

         Vertical component at maximum height = 0

So velocity at maximum height = 100 m/s

moving ball hits still ball. mover stops, hit ball moves. momentum transfer.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form