A boy tosses a ball straight up with an initial velocity of 13.0m/s a)how long does the ball take to reach its highest point? B)how high does the ball rise above it's realise point
Given,
Initial velocity of the ball (u)=13 m/s
Let the time taken by the ball to reach to highest point be t and the height reached by the ball be H.
At the top most point, the velocity of the ball will be zero, so "v=0"
Gravitational acceleration of the ball "(g)=9.8 m\/s^2"
Now, applying the first law of motion,
"v=u-gt"
Now, substituting the values,
"0=13-9.8t"
"\\Rightarrow t=\\frac{13}{9.8}sec"
Now, applying the third equation of motion,
"v^2=u^2-2gh"
Substituting the values,
"0=13^2-2\\times 9.8\\times h"
"\\Rightarrow h = \\frac{13\\times 13}{2\\times 9.8}m"
"h=8.62m"
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