a cricket ball is thrown at a speed of 28m|s in a direction 40° above the ground. calculate time of flight and the maximum height
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Answer:
maximum height = (V sin0)² / 2g
= (28 sin30)²/ 2 x 9.8
= 14 x 14 / 2 x 9.8
= 10m
time taken to return to the same level
=2 v sin0/ g
= 2 x 28 x sin30/ 9.8
= 2 x 28 x 1/ 2 x 9.8
= 2.9 sec
distance from thrower to where the ball returns
= v² sin 2Ф/ g
= 28 x 28 x sin60/ 9.8
= 69.3 m
Explanation:
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