h = 0.5*g*T^2 = 36.4,
0.5*9.8*T^2 = 36.4,
T = 2.73 s.
Vx*T = 13.16.
Vx*2.73 = 13.16,
Vx =
Cliff divers at Acapulco jump into the sea
from a cliff 36.4 m high. At the level of the
sea, a rock sticks out a horizontal distance of
13.16 m.
The acceleration of gravity is 9.8 m/s
2
.
With what minimum horizontal velocity
must the cliff divers leave the top of the cliff if
they are to miss the rock?
Answer in units of m/s
1 answer
1 answer