Question

A ball is projected at an angle of elevation of 60 ° with an initial velocity of 120m/s.calculate

1) The time taken to get to the maximum height
ii) the time of flight ​

Answer 1

Step-by-step explanation:

It is given that,

The angle of projection is 60 degrees

Initial velocity of the ball is 120 m/s

We need to find the time taken to get to the maximum height and the time of flight.

Time taken to reach the maximum height is given by :

`T=(u^2\sin^2\theta)/(2g)`

g is acceleration due to gravity

`T=((120)^2* \sin^2(60))/(2* 10)\\\\T=540\ s`

(ii) Time of flight,

`t=(2u\sin\theta)/(g)`

So,

`t=(2* 120* \sin(60))/(10)\\\\t=20.78\ s`

Hence, this is the required solution.