Question

two balls are protected simultaneously from the top of a tower, one vertically upward and other vertically downwards with the same speed of 60 m persecond then find time interval between the balls striking the ground.
​

Answer 1

Answer:

6 sec

Explanation:

as the equation for ball going downward: S1 = ut - 1/2gt²....................(1)

and the ball going vertically upward cover distance in two parts, first upward till the velocity becomes zero whose equation is:

S2 = ut - 1/2gt²...............(2)

and the equation for distance the ball cover from the velocity zero to the ground is:

v² = u² - 2gh

0² = (60)² - 2 × 10 × h

h = 180

so now as we see that the value of s1 and s2 is same as well as the initial velocity is also same so the time taken will also same. so what the difference between the time taken by both the ball will remain only by the ball 2 as to cover it the distance of 180m.

s = 1/2gt²

180 = 1/2 × 5 t²

t = 6 sec

Answer 2

Answer:

12 second.

Explanation:

According to the questions,

We know that,

The convention be as upward +ve and downward -ve

For the ball vertically upward:-

Displacement= S (let the height of the building be s)

Initial velocity, u = 60 m/s

Acceleration,a = -10 m/s ^2

Time is taken to reach the ground= t seconds

Plugging all the above values in the equation S=ut+1/2(at^2)

we get,

$-S= 60t-5t^2$ —(1)

For the ball vertically downwards:-

Displacement= -s

Initial velocity= -60 m/s

Acceleration= -g= -10m/s^2

Time is taken to reach the ground = t* seconds

So here the equation is

$-S= -60t* -5t* ^2$———-(2)

Equating (1) and (2):-

$60 t - 5 t ^2= -60 t* - 5t*^2\\\\-60 ( t + t’) +5 ( t ^2 -t'^2) =0\\\\-60( t + t*) + 5( t + t*)(t - t*)=0\\\\t - t* = 60/5 = 12 seconds\\\\t - t*= 12 seconds$

So the required time gap is 12 s.