Question

ball thrown up vertically returns to the thrower after 12s. Find; i) The velocity with which it was thrown up ii) The maximum height it reaches iii)Its position after 8s.​

Answer 1

Answer:

a) 58.8m/sec

b) 176.4m

c) 53.9m

Explanation:

a) v=u+gt

u=v-gt

0-(9.8)(6)

=58.8m/sec

b).h=v²-u²/2g

h=0-(58.8)²/2(9.8)

h=58.8×58.8/19.6

=176.4m

c) h=ut+1/2gt²

h= 0+1/2×9.8×(1)²

1/2×9.8×1

4.9m

58.8-4.9=53.9m