181k views
4 votes
A ball is thrown upward from the ground with an initial speed of 25 m/s. At the same instant, another ball is dropped from a building 15 m high. After how long will the balls be at the same height? Neglect air resistance.

a) 2 seconds
b) 3 seconds
c) 4 seconds
d) 5 seconds

User Eishay
by
8.4k points

1 Answer

5 votes

Final answer:

By using kinematic equations for the upward and downward motions of the balls, we find that the balls will be at the same height after approximately 1 second, so options (a) through (d) provided in the question are incorrect.

Step-by-step explanation:

The question asks at what time two balls, one thrown upward from the ground and the other dropped from a building, will be at the same height when taking into consideration gravitational force and neglecting air resistance. To solve this, we can use kinematic equations for projectile motion.

Let's denote the time after which they will be at the same height as 't' seconds. The upward motion can be described by the equation h = ut - 1/2 gt², where 'h' is the height, 'u' is the initial velocity, and 'g' is the acceleration due to gravity (9.8 m/s²).

The downward motion can be described by the equation h' = h - 1/2 gt². By equating both heights h = h', we set the initial velocity 'u' of the upward motion to 25 m/s, and the height 'h' from which the second ball was dropped to 15 m:

  1. For the upward motion: h = 25t - 1/2(9.8)t²
  2. For the downward motion: h' = 15 - 1/2(9.8)t²

By setting h = h' and solving for 't', we find that the balls will be at the same height after approximately 1 second. Thus, none of the options (a) 2 seconds, (b) 3 seconds, (c) 4 seconds, or (d) 5 seconds given in the question are correct.

User Sriga
by
9.2k points