An inquisitive physics student and mountain climber climbs a 48.0 m cliff that o
ID: 1653928 • Letter: A
Question
An inquisitive physics student and mountain climber climbs a 48.0 m cliff that overhangs a calm pool of water. He throws (not drops) two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 1.90 m/s downward. (a) How long after release of the first stone do the two stones hit the water? ______ (b) What initial velocity must the second stone have if they are to hit simultaneously? ______ (downward) (c) What is the speed of each stone at the instant the two hit the water? first stone ______ (downward) second stone _______ (downward)Explanation / Answer
a) 48 = 1.90 (t)+ 0.5 (9.8) t^2
4.9 t^2 + 1.90t -48=0
t = -1.90 + - sqroot ( 944.41) / 2x 4.9 = -1.90 +- 30.73/ 9,.8 = 2.94 seconds apprx
b) 48 = u ( 1.94) + 0.5 (9.8) (1.94)^2
u= 15.23 m/s apprx
c) first stone = v = 1.90 + 9.8 ( 2.94) = 30.712 m/s apprx
second stone= 15.23 + 9.8 ( 1.94)= 34.242 m/s apprx