A diver springs upward with an initial speed of 2.93 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?

at the maximum height, the PEnergy gained is equal to the initial KE

mgh=1/2 m (2.93)^2
solve for h (which is above the board).
a) for final velocity
initial PE=final KE
mg(h+3)=1/2 m vf^2
solve for vf, final velocity

To solve this problem, we need to use the equations of motion for projectile motion. Let's break it down step by step:

(a) First, let's find the time it takes for the diver to reach the water. We can use the equation for displacement in vertical motion:

y = (vi)(t) + (1/2)(a)(t^2)

Where:
- y is the displacement (in this case, y = -3.0 m)
- vi is the initial velocity (2.93 m/s)
- a is the acceleration (in this case, due to gravity, a = -9.8 m/s^2)
- t is the time we're trying to find

Rearranging the equation, we get:

(1/2)(a)(t^2) + (vi)(t) + y = 0

Substituting the given values, we have:

(1/2)(-9.8)(t^2) + (2.93)(t) - 3.0 = 0

This equation is a quadratic equation in terms of t. We can solve it using the quadratic formula:

t = (-b ± √(b^2 - 4ac))/(2a)

In our equation, a = (1/2)(-9.8), b = 2.93, and c = -3.0. Plugging these values into the quadratic formula, we get:

t = [ -2.93 ± √((2.93)^2 - 4(1/2)(-9.8)(-3.0)) ] / [2(1/2)(-9.8)]

Simplifying the equation further, we get:

t = [ -2.93 ± √(8.5721) ] / (-9.8)

Now, we can calculate the two possible values for t:

t ≈ 0.50 s or t ≈ 1.84 s

Since we're interested in the time it takes for the diver to reach the water, we choose the positive value:

t ≈ 1.84 s

Now, let's find the velocity with which the diver strikes the water. We can use the equation:

vf = vi + (a)(t)

Where:
- vf is the final velocity (the velocity with which the diver strikes the water)
- vi is the initial velocity (2.93 m/s)
- a is the acceleration (in this case, due to gravity, a = -9.8 m/s^2)
- t is the time (1.84 s)

Plugging in the given values, we have:

vf = 2.93 + (-9.8)(1.84)

Calculating this, we get:

vf ≈ -14.81 m/s

Since the downward direction is chosen as negative, the velocity with which the diver strikes the water is approximately 14.81 m/s in the downward direction.

(b) Now let's find the highest point the diver reaches above the water. The highest point occurs when the vertical velocity of the diver is zero.

We can calculate the time taken to reach the highest point using the equation:

vf = vi + (a)(t)

At the highest point, vf = 0 m/s. The initial velocity (vi) is given as 2.93 m/s, and the acceleration (a) is -9.8 m/s^2 (due to gravity). Using these values, we can solve for t:

0 = 2.93 + (-9.8)(t)

Simplifying the equation, we get:

9.8t = 2.93

t ≈ 0.30 s

To find the highest point reached, we substitute this time (0.30 s) into the equation for displacement:

y = (vi)(t) + (1/2)(a)(t^2)

For y to be positive, we consider the magnitude of the displacement:

y = |(vi)(t) + (1/2)(a)(t^2)|

Plugging in the values, we get:

y = |(2.93)(0.30) + (1/2)(-9.8)(0.30^2)|

Calculating this, we get:

y ≈ 0.44 m

Therefore, the highest point the diver reaches above the water is approximately 0.44 m.