Question

28. A ball thrown up vertically returns to the thrower after 6 second. Find
a. The velocity with which it was thrown up.
b. The maximum height it reaches, and
c. Its position after 4 second.
20. A
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Answer 1

Explanation:

à. v-0 g=9.8 (upward)

total time taken = 6 sec(up+down)

time for upward journey= 6/2= 3 sec

v= u+at

0=u+(-9.8)3

u= 29.4ms/1

b . maximum height

S= ut +1/2 a(t)^2

= (29.4*3) +1/2(-9.8)(3)^2

=88.2-44.1

h= 44.1m

c . position after 4 sec

s =ut+1/2 a(t)^2

= (29.4*4)+1/2 (-9.8)(4)^2

=117.6+(-156.8/2)

=117.6-78.4

s= 39.2m

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.