Question

Scarlet made a dive off a 10ft platform. What is the expression for the height of her heels above the surface of the pool "t" seconds after leaving the platform?

A) 10t - 16t^2
B) 10 - 16t
C) 10t + 16t^2
D) 10 + 16t

Answer 1

A) ( 10 - 16t^2 )

Step-by-step explanation:

The expression for the height of Scarlet's heels above the surface of the pool "t" seconds after leaving the platform can be modeled using the kinematic equation for free fall:

\[ h(t) = h_0 + v_0t - \frac{1}{2}gt^2 \]

In this case:

- \( h_0 \) is the initial height (10 ft, as she starts from a 10ft platform),

- \( v_0 \) is the initial velocity (0, as she starts from rest),

- \( g \) is the acceleration due to gravity (approximately 32 ft/s²).

So, the expression becomes:

\[ h(t) = 10 - \frac{1}{2}(32)t^2 \]

Simplifying this, we get:

\[ h(t) = 10 - 16t^2 \]

Therefore, the correct expression is:

A) \( 10 - 16t^2 \)

Answer 2

The correct expression for Scarlet's height above the surface of the pool t seconds after leaving a 10ft platform, accounting for gravity, is Option B: 10 - 16t^2.

Step-by-step explanation:

The expression for the height of Scarlet's heels above the surface of the pool t seconds after leaving the platform, assuming the acceleration due to gravity is approximated as 32 ft/s2 and directed downward, is given by the equation h = h0 + v0t - ½gt2, where:

• h is the height above the surface of the pool at time t,
• h0 is the initial height above the pool (the height of the platform), which is 10 ft,
• v0 is the initial velocity, which we can consider to be zero since she's just starting to fall,
• g is the acceleration due to gravity, which is 32 ft/s2 (but since it's directed downward, it should be -32 ft/s2 in the equation).

Since Scarlet is initially at rest, the v0t term is zero. Substituting h0 = 10 ft and g = -32 ft/s2 into the equation, we get:

h = 10 - 16t2, where 16 is half of 32.

Therefore, the correct expression is Option B: 10 - 16t2.