a. h = 0.5*g*T^2 = 52.
4.9*T^2 = 52,
T = 1.47 s.
b. V^2 = Vo^2 + 2g*d = 0 +29.6*52 = 1539,
V = 39. m/s.
c. Vx = 18 m/s.
d. Vt = sqrt(18^2 + 39^2) =
A person standing at the edge of a seaside cliff kicks a stone, horizontally over the edge with a velocity of 18 m/s. The cliff is 52 m above the waters surface
A. How long does it take for the stone to fall to the water
B. What is the vertical velocity component of the stone just before it hits the water
C. What is the horizontal velocity component of the stone just before it hits the water
D. What is the total velocity of the stone just before it hits the water ?
1 answer
1 answer