free fall motion

Question

2 engineering students are exploring motion of body along a straight line using calculus-based physics performing a lot of experiments. One of the students, is at the 5th floor of the building, while the 2nd student is at the 2nd floor. They throw balls (from their positions which are vertically aligned) with the same speed of 40 ft/s at the same time. If the 1st student throws the ball downward while the 2nd student throws the ball upward, calculate the speed and distance traveled by the ball thrown by the 1st student when the ball thrown by the 2nd student reaches maximum height. Use g=32ft/s^2

Answer 1

Hi Rob A.,

Using positive up,15 feet between floors, a = g = -32ft/s² (negative downward direction),

and the equation y = y₀ + v₀t + at²/2:

1st student: y₁ = 75 - 40t - 16t²; (v₀ is in the negative y-direction)

2nd student: y₂ = 30 + 40t - 16t²; (v₀ is in the positive y-direction)

Maximum height of 2nd students ball: dy₂/dt = 0

dy₂/dt = 40 - 32t =0 or t = 40/32 = 5/4

Distance traveled of 1st students ball at t = 5/4:

y₁ = 75 - 40(5/4) - 16(5/4)² = 75 - 50 - 25 = 0; The ball started at 75ft and is now at 0ft; 75ft traveled.

The speed at t = 5/4 of dy₁/dt = v₁:

v₁(t) = v₁(5/4) = dy₁/dt = -40 - 32t = -40 - 32(5/4) = -80ft/s; 80ft/s in the negative (downward) direction.

Using 15ft or greater between floors yields this result.

Use less than 15ft between floors and graph the two equations to see what happens.

I hope this helped, Joe.