First slide

Vertical projection from top of a tower

Question

From the top of a tower two bodies are projected with the same initial speed of 40 ms^–1, first body vertically upwards and second body vertically downwards. A third body is freely released from the top of the tower. If their respective times of flights are T₁, T₂ and T₃ identify the correct descending order of the times of flights

Easy

A

T₁, T₂, T₃
B

T₂, T₃, T₁
C

T₂, T₁, T₃
D

T₁, T₃, T₂

Solution

The first body will go up and after a time interval of $\large \frac{{2u}} {g} = \frac{{2 \times 40}} {{10}}s = 8s$ , it will again corss the top of the tower with a velocity of u = 40m/s and then after a time T₂, it will strike the ground. $\large \therefore$ T₁ = (8 + T₂)sec. Also since the third body is released from rest, its time of flight T₃ must be greater than T₂, so option (4) is the correct option.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App