Question

a ball is projected from a point o as shown in the fig. it ll strike the ground after (g=10)
and is 1)4s

Answer 1

here you can see body is projected with speed 10m/s at an angle 30° with horizontal from 60m above the ground.
first we have to find time taken to reach the ball at maximum height and then use time taken to reach the ground from maximum height.

time taken to reach maximum height , t =usin∅/g [formula]
= 10 × sin30°/10 = 0.5 sec

body reaches maximum height , h = u²sin²∅/2g [ formula]
= (10)² × sin²30°/2×10
= 1.25 m

now, body is at height = (60 + 1.25) = 61.25m
at maximum height, speed of body have only horizontal component.e.g., ucos30° = 5√3 m/s
now, we have to use horizontal projectile concepts to find time taken to reach the from top (61.25m).
use formula, t = √{2h/g}
here, h = 61.25 m and g = 10
now, t = √{2 × 61.25/10} = √{12.25} = 4.5 sec

now, total time = 0.5 sec + 4.5 sec = 5sec
hence, option (4) is correct.

Answer 2

Answer:

open hope it helps you thank you