Asked by tom
A golf ball is dropped from rest from a height of 8.20 m. It hits the pavement, then bounces back up, rising just 5.40 m before falling back down again. A boy then catches the ball when it is 1.40 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.
Answers
Answered by
Elena
h1=g•t1²/2=>
t1=2•h1/g= 2•8.2/9.8=1.8 s.
Since the time of ascening = time of descending,
t2=2•h2/g=1.1 s.
h3=5.4-1.4 =4 m
t3=2•h3/g=2•4/9.8=0.82 s.
t=1.8+1.1+0.82=3.72 s.
t1=2•h1/g= 2•8.2/9.8=1.8 s.
Since the time of ascening = time of descending,
t2=2•h2/g=1.1 s.
h3=5.4-1.4 =4 m
t3=2•h3/g=2•4/9.8=0.82 s.
t=1.8+1.1+0.82=3.72 s.
Answered by
RUssell
1.5
Answered by
Jordan
Elena is incorrect, in that you must take the square root of each thing.
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