Stone A is thrown vertically upward with a velocity of 60 m/s at t = 0. Stone B is thrown vertically upward with a velocity of 25 m/s at t = 2 s. The magnitude of velocity (in m/s) of stone A with respect to stone B in the time interval, 2 s < t < 7 s is [g = 10 m/s2]
Answers
Given info : Stone A is thrown vertically upward with a velocity of 60 m/s at t = 0. Stone B is thrown vertically upward with a velocity of 25 m/s at t = 2 s.
To find : The magnitude of velocity (in m/s) of stone A with respect to stone B in the time interval, 2 s < t < 7 s is [g = 10 m/s2]
Solution : case 1 : stone A is thrown upward with speed 60 m/s.
Time taken to reach highest point, t = u/g = 60/10 = 6 sec
So, after 6 sec, stone A starts to fall.
And velocity of stone A during 6s to 7 s , v = 0 - 10 m/s² × 1s = -10 m/s [ negative sign indicates stone falls downwards]
Case 2 : stone B is thrown vertically upward with a velocity of 25 m/s at t = 2s
Let's find time taken to reach maximum height, t' = u'/g = 25/10 = 2.5 sec
So stone B is at maximum height at (2 + 2.5) = 4.5s
Now velocity of stone B after falling 4.5s to 7s , v' = 0 - 10m/s² × (7 - 4.5) s = -10 × 2.5 = -25 m/s
So velocity of stone A with respect to stone B = -10 -(-25) = 15 m/s
Therefore magnitude of velocity of stone A with respect to stone B is 15 m/s