Question

A diver is standing on a platform 25 feet above the pool. He jumps from the platform at an initial velocity of 10 ft/s. Use the formula h(t)=-16t^2+vt+s where his height above the water at time t seconds, v is the starting upward velocity, and s is his starting point to write an equation to model the dive

Answer 1

Answer:

(one thing to notice, the units of the 16 in the height equation should be ft/s^2, because this is an acceleration)

We know that:

v is the initial vertical velocity, and we know that it is 10ft/s

then:

v = 10ft/s.

s is the initial position, and we know that he starts on a platform 25ft above the pool, then if we define the pool as the 0 in the vertical axis, we have:

s = 25ft

Now we can write the height as:

h(t) = (-16ft/s^2)*t^2 + (10ft/s)*t + 25ft.

Now with this equation, we can find the time it takes for the diver to hit the pool.

Remember that we defined the pool as the 0 in the vertical axis, then when h(t) = 0ft the diver will enter in the pool.

With that, we can find the value of t = time, that it takes for the diver to enter the pool.

h(t) = 0 ft = (-16ft/s^2)*t^2 + (10ft/s)*t + 25ft.

Now we can solve this with the Bhaskara equation:

$t = \frac{-10m/s +. \sqrt{(10m/s)^2 - 4*(25ft)*(-16ft/s^2)} }{-16ft/s^2*2} = \frac{-10ft/s +- 41.23ft/s}{-32ft/s^2}$

We will have two solutions:

t = ((-10 + 41.23)/-32) s = -0.96s

t = (-10 - 41.23)/-32) s = 1.6 s

The negative has no physical sense, so we can discard that option.

This means that the only correct option is 1.6 seconds, then the time that it takes for the diver to enter the pool after he jumps is 1.6 seconds.