A ball is thrown straight up is caught back by the thrower after 6 sec. Calculate

1) The velocity with which the ball was thrown up,

2) The max. height it reaches and

3) The distance travelled by the ball for 2 seconds after it hits the maximum height.

DETAILS AND ASSUMPTIONS :

i) Give the answer in the form, \[answer 1 + answer 2 + answer 3\].

ii) To make the calculations easier take the value of acceleration due to gravity as, \[g = 10 m/s^2\]

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