Question

At the super bowl point toss, the ref flipped the coin nearly straight up, reaching a maximum height above his hand of 0.8 meters. (a) With what velocity did he launch the coin? (b) How long was the coin in the air at the maximum height? (c) What is the coin's total displacement when the ref catches it? (d) With what velocity did the coin strike the ground if the ref's hand was 1.0 meters above the ground when released?

Answer 1

(a) launch velocity is u

final velocity v = 0 , displacement =Hmax = 0.8 m

=> v^2 =u^2 +2*g*H

=> 0 = u^2 - 2*(9.8)*0.8

=> u = [15.68]^0.5 = 3.95 m/s

(b) v = u + at

=> 0 = 3.95 - 9.8*t

=> t = 0.403 s ...it is in the air for 0.403 sec

(c) total displacement when ref catch it = 0

but total distance travelled = 2Hmax = 1.6 m

(d) from max height point where u=0

v striking velocity at grounnd , a = g = 9.8 m/s^2 , S = 1+0.8 = 1.8 m

v^2 = u^2 + 2*a*S

=> v^2 = 0 +2*9.8*1.8 = 35.18

=> v = 5.93 m/s