Question

a ball thrown vertically up returns to the thrower after 6 seconds find it's position after 4 s​

Answer 1

Answer:

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

For upward motion,

v=u+at

∴ 0=u+(−10)×3

⟹u=30 m/s

The maximum height reached by the ball

h=ut+ 1/2 at^2

h=30×3+ 1/2(-10)+3^2

h=45 m

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.

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.