Question

Graham and Hunter are circus performers. A cable lifts Graham into the air at a constant speed of 1.5 ft/s. When Graham’s arms are 18 ft above the ground, Hunter, who is standing directly underneath Graham, throws Graham a ball as the cable continues to lift him higher. Hunter throws the ball from a position 5 ft above the ground with an initial velocity of 24 ft/s. Which system of equations can be used to model this situation?

Answer 1

Graham's height above the ground can be modeled by the equation h_g = 18 + 1.5t, while the ball's height can be modeled by h_b = -16t^2 + 24t + 5. These equations can be used to find the time at which the ball reaches the same height as Graham during his ascent.

Step-by-step explanation:

The situation described involves projectile motion in which two objects are moving in the air at the same time, affected by gravity.

For Graham being lifted, the equation for his height above the ground as a function of time (t) can be represented as:

• h_g = 18 + 1.5t

Where 18 ft is the initial height and 1.5 ft/s is the constant speed of ascent.

For the ball thrown by Hunter, the equation for the ball's height above the ground is based on the kinematic equation:

• h_b = -16t^2 + v_0t + h_0

Where -16 ft/s^2 is the acceleration due to gravity (g = 32 ft/s^2, but since the direction is upwards, we take -g/2), v_0 is the initial velocity (24 ft/s), and h_0 is the initial height (5 ft).

The system of equations that can model this situation is:

• h_g = 18 + 1.5t
• h_b = -16t^2 + 24t + 5

These equations can be solved simultaneously to find the time at which the ball reaches Graham's height.

Answer 2

The answer to the question above as to which system of equations can be used to model the situation of Graham and Hunter when a cable lifts Graham into the air at a speed of 1.5 ft/s and Hunter throws the ball from a position of 5 ft. above the ground with an initial velocity of 24 ft/s. the equation will be : H = 18t + 1.5t - 16t ^ 2
H = 5 + 24t - 16t ^ 2