Question

solve this question and the spam will be reported and his/her account will be deleted​

Answer 1

Answer:

Explanation:

Let the initial velocity = u m/s

and angle of projection = α

x and y components are as follows;-

x=ucosαt and vx=ucosα

$y = usin\alpha t - 1/2gt^{2} -(i)$

Vy =usinα - gt       -(ii)

The horizontal component of speed v  

x  is always constant. So the speed is minimum when the vertical component Vy  =0

Putting this in equation 2

usinα=2g

The ball is moving at angle 45° direction of velocity of the project at time t=1s

Then tan45°  =  1  =     $\frac{Vy}{Vx}$ = $\frac{usin\alpha - g}{ucos\alpha }$

⇒ucosα=g

tanα=2

α=$tan^{-1} 2$  

sinα= 2/√5

cosα = 1/√5

and u =√5 g

 u = √5 × 10

 u = 10√5 m/s

PLEASE MARK IT AS BRAINLIEST