Question

AbalI is thrown vertically up ward with aveloctiy of 6m/s. (a) how long doe's it takes for the baIl to reach the maximum passible height (b) how long it take for the baIl return back to it intial poistion ? (c) with what speed doe's the bIl reach at it's intial position ? (d) what is the velocity of baIl when it reach half maximum ?

Answer 1

Answer:

Given ,

u = 6 m/s

a = g = 10 m/s²

Hmax = u² / 2g

          = 36 / 20

          = 1.8 m

a ) time taken to reach Hmax

t = v-u / a

 = -6 / -10

 = 6/10

 = 3 / 5 sec

b )time taken to return to its original position

3 / 5 + t

t = v / g

 = v / 10

v = √2×10×36 / 20

   = √36

   = 6 m/s

t = v/g

 = 6 / 10

 = 3 / 5 sec

t + 3/5

= 3/5 + 3/5

= 6 / 5 sec

c ) ball reaches its original position with speed 6 m/s as solved above

d ) velocity of ball at Hmax / 2

Hmax / 2 = 1.8 / 2

               = 0.9 m

v = √6² + 2×10×0.9

  = √36 + 18

  = √54  m/s

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