Question

if the projectile reached at the peak point of trajectory then the height attained by it 14m from the ground surface then find out the time of the ascent if the angle made by the direction of the velocity with the vertical is 60°​

Answer 1

Let the initial velocity be U,

then Ux=Ucos(60)=U/2 and Uy=Usin(60) = √3U/2

Hmax =Uy²/2g

=> 14= (√3U/2)²/2g

=> U= 1120/3

=> Uy= (1120/3)√3/2

Time of flight=2Uy/g