Question

A ball thrown up vertically returns to the thrower after 6s find
a . the velocity with which it was thrown up
b. the maximum height it reaches
c. its position after 4 s ​

Answer 1

Answer:

The ball thrown Akbar tikli returns to the thrower in 6 s , this means it take half time so

6/2 = 3

a . v = 0

u=?

g = -9.8 m/s^2 (ball goes up)

t = 3s

v= u+gt

0= u +(-9.8)×3

0= u - 29.4

u= 24.4

24.4m/s

b. v^2 = u^2+2gh

0^2=(24.4)^2+ 2(-9.8)×h

0 = 864.36-19.6h

h = 864.36/19.6

h= 44.1

Thus, the maximum height reached by the ball is 44.1m

c.

after 4 s

h= ut+1/2gt^2

h= 0×1+1/2×9.8×1^2

h= 0+ 4.9×1

h= 4.8

Thus, the position of the ball afte 4s of throwing is 4.9

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.