If Bill threw a ball straight up on the Moon (g= 1.6 m/s2) with a starting velocity of 22 m/s from a cliff and it fell past him and reached a velocity of 56 m/s as it impacted the Moon's surface below the cliff, how long was the flight time?
Answers
Given info : If Bill threw a ball straight up on the Moon (g= 1.6 m/s2) with a starting velocity of 22 m/s from a cliff and it fell past him and reached a velocity of 56 m/s as it impacted the Moon's surface below the cliff.
To find : the time of flight of ball is ..
solution : initial velocity of ball, u = 22 m/s
acceleration acting on the ball, a = 1.6 m/s²
at a point, its velocity becomes zero and then it starts to fall downwards.
time when its velocity becomes zero, t = u/a = 22/1.6 = 13.75 sec
now ball starts to fall and reaches the moon's surface below the cliff.
final speed of ball before striking the surface, v = 56 m/s
using formula, v = u + at
⇒ 56 m/s = 0 + 1.6 m/s² × t'
⇒ t' = 56/1.6 = 35 sec
∴ the time of flight = total time taken by ball to reach the moon's surface. = t + t' = 13.75 + 35 = 48.75 sec
therefore the time of flight of ball is 48.75 sec.