Question

A body is thrown vertically up from the top of the building with a velocity of 10m/s . It reaches ground after 5 seconds. Find the height of the building and the velocity with which it reaches the ground ,where g=10m/second square

Answer 1

This is your answer.... I've explained everything in it.... Hope you'll understand...

Answer 2

Given:

Velocity with which ball is thrown up from the top of building = 10 m/s

Time of flight = 5 seconds

Acceleration due to gravity g = 10 m/s²

To find:

1. Height of building
2. Velocity before touching the ground.

Solution:

Let the height of the building (AB) be h and the height to which the ball is thrown (BC) be x.

Now,

Step 1

When ball is thrown from the building . (B to C)

We have,

u = 10 m/s ; V = 0 ; a = -9.8 m/s²  ; s = x

Therefore,

$v^{2} - u^{2} = 2as$

Substituting the value, we get

$(0)^{2}-(10)^{2}= -2(10)x\\x=5m$

Hence, the ball goes 5 m above the building.

Step 2

Time taken for distance 5 m to travel will be

$s = ut+\frac{1}{2}at^{2}$

Substituting the values, we get

$5 = 10(t)+\frac{1}{2}(-10)t^{2} ; or\\t = 1s$

Hence, time taken to reach x= 5 m will be 1 sec.

Step 3

Now, given time of flight was 5 s , out of which 1 s went for travelling B to C remaining 5 - 1= 4 seconds.

Hence, the ball reaches from the top most position to the ground in 4 seconds .

We have,

s = (h + 5)m ; a = +10 m/s² ; u = 0 m/s ; t = 4 s

Substituting in given equation, we get

$(h+5)=0(4)+\frac{1}{2}(10)(4)^{2}\\h+5 = 80\\h = 75 m$

Hence, the height of the building AB is h = 75 m

Now,

Step 4

For velocity of the ball on reaching the ground.

We have,

u = 0 m/s ; a = +10 m/s²  ; s = 80 m (From C to A)

$v^{2}-u^{2}=2as\\v^{2}-(0)^{2}= 2(10)(80)\\v = 40m/s$

Final answer:

Hence, the height of the building is 75 m and the velocity of the ball with which it reaches the ground is 40 m/s.