Question

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

Answer 1

Explanation:

(a) the velocity with which thrown up

here, time of ascent =time of decent

t=6/2=3 sec

given,

v=0

a=-10m/s^2

t=3 sec

so,

v=u+at

0=u+(-10)×3

0=u-30

u-30=0

u=30m/s

(b)maximum height

v=0

u=30m/s

t=3 sec

s=h

v^2=u^2+2 as

0^2=30^2+2×10×h

0=900+20h

900+20h=0

20h=900

h=900/20

h=45m

(c)its position after 4s. here at 3s ball reached at maximum height then after 3s it start falling down. we know about position of ball in 3s.

so,

t =1 sec

u=0

a=10m/s^2

s=ut+1/2at^2

s=0×1+1/2×10×1^2

s =5m

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.