A high diver dives off a platform and strikes the water 1.857 seconds later. Ignoring air resistance, how fast is he traveling as he strikes the water in m/s?

The (vertical) speed of the driver as he strikes the water is approximately 18.22 m/s

Explanation:

From the parameters given in the question, we have;

The time after which the driver strikes the water after he drives off a platform, t = 1.857 s

Let 'v' represent the vertical speed of the driver

We can find the vertical speed of the vehicle as it strikes the water as follows;

v = u + g·t

Where;

v = The final vertical velocity of the driver

u = The initial vertical speed of the driver = 0 m/s

g = The acceleration due to gravity ≈ 9.81 m/s²

t = The time of motion = 1.857 seconds

Therefore, we have;

v = 0 + 9.81 m/s² × 1.857 s = 18.21717 m/s

Therefore;

The vertical speed of the driver as he strikes the water, v ≈ 18.22 m/s

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.