Question

4
A cricket ball is thrown at a speed of 28m/s
with a direction. 30m above horizontal
find
a) Maximum night
b) Time taken to return to same level.
c) distance from the point of projection
to a point where the ball returns to the
same level.

Answer 1

Answer:

Hii mate..

Explanation:

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..

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