Question

a ball dropped from the top of tower falls first half height of tower in 10sec the total time spend by ball in air is (take g =10m/sec^2) (a)14.14sec (b)15.26sec (c)12.36sec (d)17.36sec​

Answer 1

Given that, a ball dropped from the top of the tower falls first half-height of the tower in 10sec.

Here, time (t) = 10 sec and initial velocity of the ball (u) = 0 m/s

Also given that, a = g = 10 m/s²

Using Second Equation of Motion

s= ut + 1/2 at²

s= 0(10) + 1/2 × 10 × (10)²

s= 0 + 5 × 100

s= 500 m

First half height (s) = 500

s/2 = 500

s= 500 × 2

s= 1000 m

We have to find the total time spend by ball in air.

Again, using Second Equation of Motion

s= ut + 1/2 at²

1000 = 0(t) + 1/2 × 10 × t²

1000 = 0 + 5t²

1000 = 5t²

200 = t²

√200 = t

14.14 = t

Option a) 14.14 sec

Answer 2

Given :-

A ball dropped from the top of the tower falls first half-height of the tower in 10 s.

To find :-

The total time spent by the ball in air.

Solution :-

We know,

$\displaystyle{\sf{s=ut+\frac{1}{2} gt^2}}$

Here, g is the acceleration due to gravity. (g = 10 m/s²)

Initial velocity(u) = 0 m/s

Putting the known values in the equation :-

$\displaystyle{\sf{\implies s=0(10)+\frac{1}{2}*10*(10)^2 }}$

$\displaystyle{\sf{\implies s=0+5*100}}\\\\\displaystyle{\sf{\implies s=500\ m}}$

So, the first half-height is 500 m.

Total height = 500*2 = 1000 m

Again by using the equation,

$\displaystyle{\sf{s=0(t)+\frac{1}{2}*10*t^2 }}\\\\\displaystyle{\sf{\implies 1000=0+5*(t)^2}}\\\\\displaystyle{\sf{\implies 1000=5t^2}}\\\\\displaystyle{\sf{\implies t^2=200}}\\\\\boxed{\displaystyle{\implies \sf{t=14.14\ s}}}$

So, the answer is (a) 14,14 sec.