Two balls are thrown simultaneously, ball(1) downwards from the top, and ball(2)
ID: 2172362 • Letter: T
Question
Two balls are thrown simultaneously, ball(1) downwards from the top, and ball(2) upwards from the bottom, of a building of height 80.0 m, both with initial velocities of 20.0 m/s. Use a coordinate system with y = 0 at ground level, positive y direction pointing upwards, and acceleration due to gravity, and g = 9.80 m/s2.a) (2 points) Express the positions, y1 and y2, of the two balls as a function of time? y1 = _______________
y2 =_______________
b) (2 points) How many seconds after release will they meet (have the same y value)? ________________
c) (1 point) At what height above ground will they meet?
Explanation / Answer
from the base of the cliff: s = vt + ½at² = 9m/s•t - 4.9m/s²•t² from the top: s = 6m - 9m/s•t - 4.9m/s² Then meet where s = s: 9t = 6 - 9t 18t = 6 t = 0.33s bottom: s = 9•0.33 - 4.9•0.33² = 2.45m top: s = 6 - 9•0.33 - 4.9•0.33² = 2.45m v 2) Basic kinetmatics stuff. Use s = ½at² to find the time to fall s = 9.5m Then use kinematics to find the time to fall 5.7m (It's actually rising, but time to fall = time to rise, and fall time is easier to calculate (no initial v)!) Then use kinematics to find the time to fall (5.7m - 1.2m). Add the three times for your answer.