A boy jumps off a tower 1000 feet high with an initial velocity of 120 ft/sec. At the same time, a girl walks on the ground along the foot of the tower

Question

according to the equation of motion G(𝑡) = 𝑡^3 −10𝑡^2 + 25𝑡 + 1, where G is the girl’s directed distance at any time 𝑡 from the point directly under the path of the boy’s path of motion.

1. What is the boy's equation of motion?
2. What is the boy's speed when he hits the ground?
3. At what time is the girl accelerating?
4. Upon hitting the ground, will the boy hit the girl? Explain.

Answer 1

g= -32 ft/s² v = -120 ft/s

1) B(t) = gt²/2 + v₀t + h = -16t²-120t+1000

2)

When B(t) = 0 t = (120±√(120²-4(-16)(1000))/ -32 = 5, -12.5 Use 5

v(t)= B'(t) = -32t -120

v(5) = -280 ft/s or 280 ft/s down

3) The critical points for G(t) is when G'(t) = 3t²-20t +25 = 0: t = 1.67, 5

G(1.67) is a maximum and G(5) is a minimum which can be determined by the first or second derivative

tests or by looking at the plot of G(t).

The girl is accelerating [0,1.67] and again at [5, ∞)

4) No, but close. G(5) = 1 not zero