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A ball is threown straight upward with a speed v from a point h meters above the ground.Show that the time taken for the ball to strike the ground is (v/g)[1+ sqrt{1+(2hg/v^2)}].

DEBASISH , 6 Years ago

Grade 11

1 Answers

Eshan

Last Activity: 6 Years ago

Dear student,

Taking the vertically upwards direction as positive, for the entire motion, the displament of the ball is $s=-h$ . Initial velocity of ball is $u=+v$ and acceleration is $a=-g$

Since $s=ut+\dfrac{1}{2}at^2$

$\implies -h=vt-\dfrac{1}{2}gt^2$
$\implies t^2-\dfrac{2v}{g}t-\dfrac{2h}{g}=0$
$\implies t=\dfrac{v}{g}+\sqrt{\dfrac{v^2}{g^2}+\dfrac{2h}{g}} =\dfrac{v}{g}(1+\sqrt{1+\dfrac{2gh}{v^2}})$

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