A70.0-kg ice hockey goalie, originally at rest, has a 0.280-kg hockey puck slapped at him at a velocity of 45.5 m/s. suppose the goalie and the puck have an elastic collision, and the puck is reflected back in the direction from which it came. what would the final velocities of the goalie and the puck be in this case? assume that the collision is completely elastic.

mₐ vᵢₐ + mᵦ vᵢᵦ = mₐ vfₐ + mᵦ vfᵦ ==> [i and f means initial and final velocities] 

0.5 mₐ vᵢₐ² + 0.5 mᵦ vᵢᵦ² = 0.5 mₐ vfₐ² + 0.5 mᵦ vfᵦ² 

The collision is completely specified given the two initial velocities and the masses of the two objects. Combining the two equations above gives us a solution to the final velocities for an elastic collision of two objects: 

vfₐ = [(mₐ - mᵦ) vᵢₐ + 2 mᵦ vᵢᵦ]/[mₐ + mᵦ] 

vfᵦ = [2 mₐ vᵢₐ − (mₐ - mᵦ) vᵢᵦ]/[mₐ + mᵦ] 

substituting the values 

mₐ = 70 kg 
mᵦ = 0.280 kg 
vᵢₐ = 0 m/s 
vᵢᵦ = 45.5 m/s 

=> vfₐ = [(70 - 0.280) 0 + 2(0.280) 45.5]/[70 + 0.28] 

= 1.35 m/s >(in the original direction of the puck) 

and 

=> vfᵦ = [0 − (70 - 0.280) 45.5]/[ 70 + 0.28] = -45.14 m/s (opposite direction) 

Final speeds are therefore:

goalie = 1.35 m/s

puck = 45.14 m/s (opposite direction)

answer: what are you talking about?

explanation:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft