A driver jumps off the 3.0m diving board with an initial velocity of 5.0m/s. (a) Determine the maximum height, (b) The time to reach the maximum height, (c) The time it took to touch the water, and (c) The speed when touching the water.

assuming the initial velocity is nearly straight upwards...

at max height, velocity is zero
0=5-9.8t
time to max height= 5/9.8 seconds

time to water:
hf=hi+vi*t-4.8t^2
0=3+5*t-4.9t^2
solve this quadratic equation for t.

speed:
vf=5*t-9.8t^2 put in t, and you have it.

To solve this problem, we will use the equations of motion for vertical motion under constant acceleration.

Given:
Initial velocity (u) = 5.0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height of the diving board (h) = 3.0 m

(a) Maximum Height:
The maximum height is reached when the velocity becomes zero. We can use the equation: v² = u² + 2gh
where v is the final velocity (0 m/s) and h is the maximum height. Rearranging the equation, we get: h = (v² - u²) / (2g)

Substituting the given values:
h = (0² - 5.0²) / (2 * 9.8)
h = (-25.0) / 19.6
h ≈ -1.28 m

Since we obtain a negative value for height, it indicates that the maximum height is below the starting point. Therefore, the driver cannot reach the maximum height.

(b) Time to reach the maximum height:
Again, we can use the equation: v = u + gt, where v is the final velocity (0 m/s) and t is the time taken to reach the maximum height.
Rearranging the equation, we get: t = (v - u) / g

Substituting the given values:
t = (0 - 5.0) / 9.8
t = -5.0 / 9.8
t ≈ -0.51 s

Since we obtain a negative time, it indicates that the driver took 0.51 seconds to reach the starting point. This result is not meaningful, as time cannot be negative.

(c) Time to touch the water:
To answer this question, we need to find the time taken to fall from the diving board to the water. We can use the equation: h = ut + (1/2)gt², where h is the height of the diving board (3.0 m), u is the initial velocity (5.0 m/s), g is the acceleration due to gravity (9.8 m/s²), and t is the time.

Rearranging the equation, we get: t² + (2u / g) * t - (2h / g) = 0

Substituting the given values:
t² + (2 * 5.0 / 9.8) * t - (2 * 3.0 / 9.8) = 0

Solving this quadratic equation, we find the values of t. The positive value will represent the time taken to touch the water.

Using the quadratic formula: t = (-b ± √(b² - 4ac)) / (2a)

t = [-(2 * 5.0/9.8) ± √((2 * 5.0/9.8)² - 4(1)(-2 * 3.0/9.8)) / (2 * 1)]

Simplifying the equation:
t = [-1.02 ± √(1.0404 + 0.2449)] / 2
t = [-1.02 ± √1.2853] / 2

Solving for t:
t ≈ [-1.02 + 1.13] / 2 ≈ 0.053 s (Ignoring the negative value)

The time taken to touch the water is approximately 0.053 seconds.

(d) Speed when touching the water:
To find the speed when touching the water, we can use the equation: v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken to touch the water.

Substituting the given values:
v = 5.0 + (9.8 * 0.053)
v ≈ 5.51 m/s

The speed when touching the water is approximately 5.51 m/s.

To solve this problem, we can use the equations of motion for an object in free fall. Let's go through each part of the question step by step.

(a) Determine the maximum height:
The maximum height is reached when the driver's vertical velocity becomes zero. We can use the equation for vertical displacement to find this height. The equation is:
h = (v^2 - u^2) / (2g)

Where:
h = maximum height
v = final vertical velocity (which is 0 m/s at the maximum height)
u = initial vertical velocity (5.0 m/s)
g = acceleration due to gravity (9.8 m/s^2)

Substituting the given values into the equation:
h = (0^2 - 5.0^2) / (2 * 9.8)
h = (-25) / 19.6
h ≈ -1.275 m

Since the height cannot be negative, the maximum height is approximately 1.275 meters above the diving board.

(b) The time to reach the maximum height:
To find the time taken to reach the maximum height, we can use the equation for vertical velocity. It is given by:
v = u + gt

Where:
v = final vertical velocity (0 m/s at the maximum height)
u = initial vertical velocity (5.0 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time

Substituting the given values:
0 = 5.0 + (9.8)t
-5.0 = 9.8t
t = -5.0 / 9.8
t ≈ -0.51 seconds

Again, time cannot be negative, so we disregard the negative sign. The time taken to reach the maximum height is approximately 0.51 seconds.

(c) The time it took to touch the water:
Since the time taken to reach the maximum height is 0.51 seconds, the time taken to touch the water will be twice that amount. This is because the ball follows the same path while going up and coming down. Thus, the total time spent in the air is twice the time taken to reach the maximum height.

Therefore, the time it took to touch the water is 2 * 0.51 = 1.02 seconds.

(d) The speed when touching the water:
To find the speed when touching the water, we can use the formula for vertical velocity. It is given by:
v = u + gt

Where:
v = final vertical velocity at the time of impact (which is what we need to find)
u = initial vertical velocity (5.0 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken to touch the water (1.02 seconds)

Substituting the given values:
v = 5.0 + (9.8)(1.02)
v ≈ 14.196 m/s

Therefore, the speed at the time of impact when touching the water is approximately 14.196 m/s.