"A diver springs upward with an initial speed of 1.8 m/s from a 3.0 m board. Find the velocity with which the diver strikes the water."
Help please!
Is initial energy (KE and PE) has to equal the KE at impact.
KEimpact=PEorig+KEorig
1/2 m vf^2=mgh + 1/2 m vi^2
solve for vf.
Notice mass divides out, you know h=3, vi=1.8m/s, and g is Earth's gravitation field (9.8N/kg)
How far above the board will he go?
v = Vi - 9.8 t
0 = 1.8 - 9.8 t
t = .184 seconds until he stops at the top
how far up above the board was that?
h = 0 + 1.8 t - 4.9 t^2
h = 1.8(.184) - 4.9(.184^2)
= .331-.165 = .166 meters above the board
so he falls 3+.166 = 3.166 meters
3.166 = 4.9 t^2
t = .804 seconds falling from top
v = 0 - 9.8 (.804)
= -7.88 meters/second
Thank you! I got it!
To find the velocity with which the diver strikes the water, we can use the principle of conservation of energy.
The initial kinetic energy of the diver is converted into potential energy as the diver moves upwards. At its highest point, all of the initial kinetic energy is converted into potential energy, given by mgh, where m is the mass of the diver, g is the acceleration due to gravity, and h is the height from which the diver started.
At this point, the potential energy is then converted back into kinetic energy as the diver falls downward towards the water. According to the law of conservation of energy, the potential energy when the diver is at the highest point is equal to the kinetic energy just before the diver strikes the water.
Mathematically, this can be represented as:
(1/2)mv^2 = mgh
where v is the velocity with which the diver strikes the water.
Since the initial speed is given as 1.8 m/s and the height from which the diver started is 3.0 m, we can substitute these values into the equation:
(1/2)(1.8^2) = 9.8(3.0)
Solving for v, we get:
v^2 = 58.8
Taking the square root of both sides of the equation, we find:
v ≈ 7.67 m/s
Therefore, the velocity with which the diver strikes the water is approximately 7.67 m/s.