Question

A ball thrown vertically up returns to the thrower after 6s. find the velocity wih which it was thrown and its position after 4s.

Answer 1

∆Velocity of projection is.....
∆Time of flight =2u/g
∆by taking g is equal to 10m/s^2
∆then,u=6×10/2=30m/s^1

∆ It's position at 4th sec is 1st sec of free fall
∆ distance travelled in 4th sec is
∆ formula is Snth= u+ g/2(2n-1)
∆ as u is equal to zero and n is equal to 1
∆ then formula become Sn= g/2 then, distance travelled is 10/2 =5m.

✓This is the answer.

Answer 2

Answer:

Explanation:

The ball returns to the ground after 6 seconds.

Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s

Let the velocity with which it is thrown up be  u

(a). For upward motion,                

v=u+at

∴     0=u+(−10)×3                  

⟹u=30  m/s      

(b). The maximum height reached by the ball

h=ut+  1 /2 at2

h=30×3+  1/2 (−10)×3  2

               

h=45 m  

(c). After 3 second, it starts to fall down.  

Let the distance by which it fall in 1 s   be   d

d=0+  1/2 at  2 ′        

where   t  =1 s

′

d=  1/2×10×(1)  2

=5 m  

∴ Its height above the ground, h  

′

=45−5=40 m  

Hence after 4 s, the ball is at a height of 40 m above the ground.