A diver jumps off a 3.00 meter high diving board with an initial velocity of 1.75 m/s vertically upward.
a) What is the diver's veolcity when she reaches the water? (assume the surface of the water is 3.00 meters below the board.)
b) What is the highest point the diver reaches above the water?
The Attempt at a Solution
a) Since we know the height, initial velocity and acceleration, we use V^2= Vo^2 + 2ad
v=square root of vo^2 + 2ad
v = square root of 1.75^2 + 2(9.80)(3)
v= 7.86527 m/s = 7.87m/s
b)we use the same equation but this time we only calculate the diver going up, the maximum height of the the jump.
V^2 = Vo^2 + 2ad
The maximum height = diver's jump + the 3.00m below the diving board.
so the answer is 3.3125m is the highest point the diver reaches above the water.
Can anyone check see if what i did is right? Thank you.