A person standing on the edge of a cliff of height 40 m above the ground throws the first ball straight up with an initial velocity of 20 ms, and then throws the second ball straight down with the same initial velocity 20 ms. What are their respective velocities just before they hit the ground? (Take a = 10 m/s2)
20√3 m/s, 20√3 m/s
For the first ball, consider when it is going upwards to reach the maximum height.
From third equation of motion,v2=u2+2as
0 = 202 - 2 × 10 × s
s = 20 m
Total distance = 20 m + 40 m = 60 m
Now, consider second part where ball from topmost point starts falling down and reaches the ground
v2=u2+2as
v2 = 0 + 2 × 10 × 60
v = √1200 = 20√3 m/s
For the second ball,
v2=u2+2as
v2 = 202 + 2 × 10 × 40 = 1200
v = 20√3 m/s