Question

A stone is dropped from top of a tower 300 m high and at the same time another is projected vertically
[
1
meet.
1) 2s, 200.9m
2) 3s, 255.9m
3) 4s, 250.8m
4) 5s, 255.10m
upward with a velocity of 100 ms-1 from the foot of the tower. Find when and where the two stones​

Answer 1

Answer:

2) 3s, 255m

Explanation:

Consider the following picture. The stone which is freely falling is named 'A' and the another stone which is vertically projected at 90° is named 'B'.

The distance after which the stones meet is taken to be 'x'

Note: acceleration due to gravity is taken to be 10 m/s

For stone 'A':

$s_{a} = ut + \frac{1}{2} g \: {t}^{2}$

$300 - x = 0 + \frac{1}{2} \times 10 \times {t}^{2}$

$300 - x = 5 {t}^{2}$

u is taken to be 0 because it freely falling.

take the above equation as (1)

where 't' is the time after which the stones meet.

For stone 'B':

$x = ut + \frac{1}{2} g {t}^{2}$

here u = 100m/s

and g= - 10m/s

$x = 100t - \frac{1}{2} \times 10 \times {t}^{2}$

$x = 100t \: - \: 5 {t}^{2}$

take the above to be (2)

substitute (2) In (1)

$300 - 100t + 5 {t}^{2} = 5 {t}^{2}$

$300 = 100t$

$t = \frac{300}{100}$

$t = 3s$

take the above equation as (3)

substitute (3) in (2)

$x = 100(3) \: - \: 5 \times ({3)}^{2}$

$x = 300 - 45$

$x = 255m$