From a 200 m high tower, one ball is thrown upwards with speed of 10 ms⁻¹ and another is thrown vertically downwards at the same speeds simultaneously. The time difference of their reaching the ground will be nearest to(a) 12 s (b) 6 s(c) 2 s (d) 1 s
Answers
The ball thrown upwards, when it comes back to the starting point, will trace out the same path as the other ball, since the following velocity and acceleration profiles are the same (both will go downward at 10m/s, with acceleration 10m/s²).
So the only difference in time caused is the extra path the ball thrown upwards traverses.
This time shall be twice of the time it will take to reach the top of the trajectory and stop.
It will be given by:
T = 2t = 2 * (v - u)/g
Or, T = 2 * (0 - (-10)) / 10 s = 2s (c).
Answer:
C) 2s
Explanation:
The total time taken by Ball-1 from t-0 to the moment it reaches ground = t-1
Let the ball = t-2
We need (t-1) - (t-2)
For, t-1
Upward phase, u = 10m/s a= -10m/s² v=0,
time = v-u/a = -10/-10 = 1 second
Height attained during the initial phase = ut + 1/2gt² = 10×1 + 1/2(-10)(1)² = 5m
Downward Phase, u = 0 height = 205 m,
= 205m = 0 + 1/2(10)t²
t = √41
t-1 = 6.4031+1 = 7.4031
For t-2
Height traveled = 200 m, u = 10m/s
200 m = 10t + 1/2(10)t²
5t²+10t-200 = 0
t² + 2t - 40 = 0
t = [-2 √4–(4)(1)(-40)]/2 = [-1(+-)(1/2 √(164)] = -1 (+-) 12.8062/2 = 5.4031
Difference = 7.4031–5.4031
= 2
Thus, the time difference of their reaching the ground will be nearest to 2 Seconds.