Upside down !

Classical Mechanics Level 2

A stone is allowed to fall from the top of the tower 100 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 25 m/s. The two stones will meet above the ground at the height of :

Details

  • Give answer in meters.
  • Take g = g= 10 m/s 2 ^2


The answer is 20.

Shivamani Patil by shivamani patil , 21, India

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1 solution

Chew-Seong Cheong
Jul 30, 2014

Let the stones meet at height h h after time t t . The distance the falling stone travels is given by: 100 − h = 1 2 g t 2 ⇒ 100 = h + 1 2 g t 2 100-h=\frac {1}{2}gt^2 \Rightarrow 100 = h + \frac {1}{2}gt^2 The distance the projected stone travels is given by: h = 25 t − 1 2 g t 2 ⇒ 25 t = h + 1 2 g t 2 h=25t-\frac {1}{2}gt^2 \Rightarrow 25t = h + \frac {1}{2}gt^2 From the two equations, we have: 25 t = 100 ⇒ t = 4 s 25t=100 \Rightarrow t = 4 s Using g = 10 g = 10 m / s m/s , when t = 4 t=4 s s , h = 25 × 4 − 0.5 × 10 × 16 = 100 − 80 = 20 m h = 25\times 4 - 0.5 \times 10 \times 16 = 100 - 80 = 20 m Similarly, 100 − h = 80 ⇒ h = 20 m 100-h=80 \Rightarrow h = \boxed {20} m

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I really admire your solutions!

Swapnil Das - 5 years, 11 months ago

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Thanks. I will continue to work on them.

Chew-Seong Cheong - 5 years, 11 months ago

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