Question 13 Marks
A ball is thrown upward and reaches a maximum height of $19.6 \ m$. Find its initial speed?
Answer
View full question & answer→Maximum Height attained by ball $= 19.6 m$
Let initial speed of ball $= u ms^{-1}.$
Acceleration applied on ball due to gravity$ = -9.8 ms^{-2}.$
Final speed of ball at maximum height $= 0 ms^{-1}.$
We know that from second equation of motion
$V^2 - u^2 = 2as$
$0 -u^2 = 2 X(-9.8)X19.6$
$u^2 = 19.6 X 19.6$
$u= 19.6 ms^{-1}$
So initial speed of ball to attain maximum height of 19.6 m should be $19.6ms^{-1}.$
Let initial speed of ball $= u ms^{-1}.$
Acceleration applied on ball due to gravity$ = -9.8 ms^{-2}.$
Final speed of ball at maximum height $= 0 ms^{-1}.$
We know that from second equation of motion
$V^2 - u^2 = 2as$
$0 -u^2 = 2 X(-9.8)X19.6$
$u^2 = 19.6 X 19.6$
$u= 19.6 ms^{-1}$
So initial speed of ball to attain maximum height of 19.6 m should be $19.6ms^{-1}.$

