Answer:
Explanation:
Given:
- initial velocity of projectile,
- angle of projection above horizontal,
height of the initial projection point above the ground,
<u>Vertical component of the velocity:</u>
<u>The time taken in course of going up:</u>
(at top the final velocity will be zero)
<u>In course of going up the maximum height reached form the initial point:</u>
(at top height the final velocity is zero. )
using eq. of motion,
where:
final vertical velocity while going up.=0
maximum height
<u>Now the total height to be descended:</u>
Now the time taken to fall the gross height in course of falling from the top:
<u>Now the total time the projectile spends in the air:</u>
<u>Now the horizontal component of the initial velocity:</u>
(it remains constant throughout the motion)
Therefore the horizontal distance covered in the total time;