Question

A ball is dropped from the top of a building of height 100 metres and after 1.05 seconds another ball is thrown vertically downwards from a window at height 70 m from the ground if both balls reach the ground at the same time find the initial velocity of the thrown ball?

Answer 1

Answer:

7.7 m/s is the correct answer

Explanation:

Mark the answer brainliests please

Answer 2

Given :

Height of 1st building :

H=100 m

Height of 2nd building :

h= 70 m

Initial speed of 1st ball :

= 0

Time after which 2nd ball is thrown :

= 1.05 sec

Both balls reach the ground at same time .

To Find :

Initial velocity of the 2nd ball or thrown ball= ?

Solution :

Time taken by the 1st ball to cover 100 m will be given as :

$H=ut +\frac{1}{2}\times gt^{2}$

Since the ball is dropped so its initial velocity will be  , so time taken will be :

$t= \sqrt{2\frac{H}{g} } \\\\t=\sqrt{2\frac{100}{10}$

$t=\sqrt{20} \\t=4.47 sec$

Now the 2nd ball is thrown from height of 70 m after 1.05 sec the 1st ball is thrown , so time it will take to reach the ground :

T=t - 1.05 sec

T = 3.42 sec

Let the initial velocity of the 2nd ball be :

= v

So using the same formula for 2nd ball also :

$h=vt + \frac{1}{2} gT^{2} \\\\70 = v\times 3.42 + \frac{1}{2} \times 10\times (3.42)^{2} \\\\70 = 3.42v + 5\times 11.7\\3.42v =70- 58.48\\v=\frac{11.52}{3.42} \\\\v=3.37 \frac{m}{s}$

So the initial velocity of thrown ball is 3.37 meter per second .