Question

An apple is dropped from top of a multistorey building. At the same instant an arrow is shot vertically upwards from the ground with velocity 11m/s . The arrow hits the apple after 2s. Calculate the height of the building. ​

Answer 1

We have been given that initial velocity ( u ) = 0 m/s . This is because it was dropped from rest .

The final velocity ( v ) = 11 m/s .

We are also given that the time taken is 2 s .

Take acceleration due to gravity ( g ) = 10 m/s² .

Height of the building = distance travelled by the apple .

Using the laws of motion :

S = ut + 1/2 at²

⇒ S = 0 + 1/2 at² [ u = 0 ]

⇒ S = 1/2 at²

Put a = 10 and t = 2

⇒ S = 1/2 × 10 × 2²

⇒ S = 1/2 × 10 × 4

⇒ S = 2 × 10

⇒ S = 20 m

The height of the building is 20 m .

Answer 2

Given,

Speed of arrow = 11m/s

Time after which the apple gets hit by arrow= 2s.

Hence, here the height of the building is given by the sum of distance travelled by arrow in 2 seconds and distance travelled by apple dropped from building in two seconds.

Refer to attachment for diagram.

Now, in case of the falling apple =>

Time(t) = 2 seconds

Initial velocity (u)= 0m/s

(since it was stationery before the fall)

Acceleration(a)=g=10m/s^2

(Since the apple was falling the accelearation will same that of accelearation due to gravity of earth)

Now, by using formulas of kinematics =>

S= ut + (1/2)at^2

=>S= 0×2 + (1/2)×(10)× (2)^2

=>S = 1/2 ×(10) × 4

=>S =(20)m. ---------------(i)

Hence,the distance travelled by the apple in 2 seconds is 20m downwards.

Now,in case of arrow=>

initial velocity(u)= 11m/s

accelearation(a)=g= -10m/s^2

time (t) = 2 seconds

By using formulas of kinematics we get=>

S=ut + 1/2a×t^2

=>S = 11(2) + 1/2 ×(-10) × (2)^2

=>S = 22 - 20

=>S = 2m. -----------(ii)

So distance travelled by the arrow in 2 seconds is 2m.

hence,the total distance will be

=>(i) + (ii)

=>20m + 2m

=>22m