Question

A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of O 100 m/s. 60 m/s. 50 m/s 80 m/s. 110 m/s.

Answer 1

The initial velocity is 50 m/s.

(C) is correct option.

Step-by-step explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

`v^2=u^2+2gh`

`h =(v^2)/(2g)`....(I)

For second half,

We need to calculate the time

Using equation of motion

`h =ut+(1)/(2)gt_(2)^2`

`h=0+(1)/(2)gt_(2)^2`

`t_(2)=\sqrt{(2h)/(g)}`

Put the value of h from equation (I)

`t_(2)=\sqrt{(2* v^2)/(g^2)}`

`t_(2)=(v)/(g)`

According to question,

`t_(1)+t_(2)=10`

`t_(1)=t_(2)`

Put the value of t₁ and t₂

`(v)/(g)+(v)/(g)=10`

`(2v)/(g)=10`

`v=(10* g)/(2)`

Here, g = 10

The initial velocity is

`v=(10*10)/(2)`

`v=50\ m/s`

Hence, The initial velocity is 50 m/s.