Question

Both Rachel and Dominique throw tennis balls into the air. At any time, t, the height, h, of Rachel’s ball is modeled by the equation h = –16t2 + 30t + 5. Dominique throws his tennis ball with the same acceleration, a, from the same initial height, h = -16t2 + 30t - 5, but with an initial velocity, v, double that of Rachel’s. Which equation best models the height of Dominique’s tennis ball?

Answer 1

Answer:

h = -16t^2+60t+5

Step-by-step explanation:

We have given that:

Rachel’s ball is modeled by the equation h = –16t2 + 30t + 5.

Dominique throws his tennis ball with the same acceleration, a, from the same initial height, h = -16t2 + 30t - 5, but with an initial velocity, v, double that of Rachel’s.

We have to find:

Which equation best models the height of Dominique’s tennis ball?

Solution:

The equation we have been given is:

h = –16t2 + 30t + 5.

where -16 is the acceleration

30 is the initial velocity

5 is the initial height

So when the initial velocity is doubled then the final equation we get is:

h = -16t^2+60t+5

Thus the equation that models the height of Dominique's tennis ball is

h = -16t^2+60t+5

Answer 2

Answer:

$h=-16t^2+60t+5$

Step-by-step explanation:

The general equation of projectile motion is

$y=\frac{1}{2}at^2+vt+h_0$            ....(1)

where, a is acceleration, v is initial velocity and $h_0=5$.

It is given that Rachel’s ball is modeled by the equation

$y=-16t^2+30t+5$              .... (2)

On comparing both sides we get

$\frac{1}{2}a=-16$

$v=30$

$h_0=5$

It is given that Dominique throws his tennis ball with the same acceleration, a, from the same initial height but with an initial velocity, v, double that of Rachel’s.  

For Dominique's model

$\frac{1}{2}a=-16$

$v=2\times 30=60$

$h_0=5$

Therefore, Dominique’s ball is modeled by the equation $h=-16t^2+60t+5$.