Question

A worker is being raised in a bucket lift at a constant speed of 3 ft/s. When the worker's arms are 10 ft off the ground, her coworker throws a measuring tape toward her. The measuring tape is thrown from a height of 6 ft with an initial vertical velocity of 15 ft/s. Projectile motion formula: h = -16t2 + vt + h0 t = time, in seconds, since the measuring tape was thrown h = height, in feet, above the ground Which system models this situation? h = 3t+10 and h = -16t2+15t+6 10t+3 and h = -16t2+6t+15 -16t2+3t + 10 and h =-16t2+15t+6 -16t2+10t + 3 and h = -16t2+6t+15

Answer 1

H=3t+10 and h=-16t+15t+6

They will never be at the same height at the same time

Answer 2

Answer:

The option 1 is correct.

Step-by-step explanation:

It is given that the worker is being raised in a bucket lift at a constant speed of 3 ft/s. When the worker's arms are 10 ft off the ground, her coworker throws a measuring tape toward her.

$h=a+bt$

Where a is initial height from the ground and b is rate of change in the height of workers hand from the ground.

$h=10+3t$

The measuring tape is thrown from a height of 6 ft with an initial vertical velocity of 15 ft/s. Projectile motion formula,

$h=-16t^2+vt+h_0$  ..... (1)

Where v is the velocity and $h_0$ is the initial height. The height of measuring tape is defined by the equation.

$h=-16t^2+15t+6$       .... (2)

Therefore the system of equation contains equation (1) and (2).

From the figure we can say that the height is not same for any value of t because the graph does not intersect each other. So we can say that the worker is not able to catch the measuring tape.