A ball is dropped from the top of tower falls first half height of tower in 10sec.The total time spent by the ball in air is?
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Answered by
107
According to the question
The half height of tower= distane travelled by ball in 10 sec
= ut+½at²
=0 + ½*9.8*10²
=490m
hence ½height of tower=490m
whole height of tower=980m
now again using for full height=980m
H=ut+½at²
or. 0 +½*9.8*t²=980
hence t²=2*980*/9.8
or t²=2*100
time of flight=10√2 sec
The half height of tower= distane travelled by ball in 10 sec
= ut+½at²
=0 + ½*9.8*10²
=490m
hence ½height of tower=490m
whole height of tower=980m
now again using for full height=980m
H=ut+½at²
or. 0 +½*9.8*t²=980
hence t²=2*980*/9.8
or t²=2*100
time of flight=10√2 sec
Answered by
72
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