A ball is dropped from a 100m high tower. Simultaneously another ball was thrown up with a velocity of 50 M per second. After how much time would these balls cross each other?
Answers
Answer:
Time duration : 2 sec
Step-by-step explanation:
Ball1: Dropped from 100 m
Its height at time t : h1 = 100 - (0 * t + 1/2 g t²) = 100 - 1/2 g t²
Ball2: Thrown up with u = 50 m/s
Its height at time t : h2 = u t - 1/2 g t² = 50 t - 1/2 g t²
They collide when h1 = h2.
=> 100 - 1/2 g t² = 50 t - 1/2 g t²
=> t = 2 sec.
At t = 2 s, h1 = h2 = 100 - 2 g or 80 m for g = 10 m/s²
Velocity of Ball1: v1 = 0 + g t = 2 g m/s downwards.
Velocity of ball2: v2 = 50 - g t = 50 - 2 g m/s upwards.
Answer:
Step-by-step explanation:
Ball1: Dropped from 100 m
Its height at time t : h1 = 100 - (0 * t + 1/2 g t²) = 100 - 1/2 g t²
Ball2: Thrown up with u = 50 m/s
Its height at time t : h2 = u t - 1/2 g t² = 50 t - 1/2 g t²
They collide when h1 = h2.
=> 100 - 1/2 g t² = 50 t - 1/2 g t²
=> t = 2 sec.
At t = 2 s, h1 = h2 = 100 - 2 g or 80 m for g = 10 m/s²
Velocity of Ball1: v1 = 0 + g t = 2 g m/s downwards.
Velocity of ball2: v2 = 50 - g t = 50 - 2 g m/s upwards