Question
A stone is dropped from a tower $98 $m high. With what speed should a second stone be thrown $1$ s later so that both hit the ground at the same time?
✓
Answer
Height of tower $=98 m$
Acceleration due to gravity on stone $=9.8 ms^{-2}$.
Initial speed of ball $=0 ms^{-1}$.
Let initial speed of second stone is $v ms ^{-1}$.
We know from second equation of motion
$S = ut + 1/2 a Xt^2.$
$98 = 0 + 1/2 X9.8Xt^2.$
$t^2 = 20$
$t= 4.47 \sec.$
As second stone is thrown $1$ sec later so time taken by second body to cover distance of $98 m$ is $= 4.47 - 1 = 3.47sec.$
So again put t= 3.47 sec and S = 98 m in second equation of motion we get
$98 = vX3.47 + 1/2 X9.8X3.47X3.47.$
$98 = 3.47Xv + 59$
$3.47X v = 98 - 59$
$v= 39/3.47 = 11.23 ms^{-1}.$
Initial speed of second stone should be $11.23 ms^{-1}.$
Acceleration due to gravity on stone $=9.8 ms^{-2}$.
Initial speed of ball $=0 ms^{-1}$.
Let initial speed of second stone is $v ms ^{-1}$.
We know from second equation of motion
$S = ut + 1/2 a Xt^2.$
$98 = 0 + 1/2 X9.8Xt^2.$
$t^2 = 20$
$t= 4.47 \sec.$
As second stone is thrown $1$ sec later so time taken by second body to cover distance of $98 m$ is $= 4.47 - 1 = 3.47sec.$
So again put t= 3.47 sec and S = 98 m in second equation of motion we get
$98 = vX3.47 + 1/2 X9.8X3.47X3.47.$
$98 = 3.47Xv + 59$
$3.47X v = 98 - 59$
$v= 39/3.47 = 11.23 ms^{-1}.$
Initial speed of second stone should be $11.23 ms^{-1}.$
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