A body is released from the top of the tower of height 100m . After 2

1.	A body is released from the top of the tower of height 100m . After 2 seconds it is stopped and then instantaneously released. what will be its height after next 3 seconds? [Take g=10m/s ^-2]1》40m2》35m3》45m4》30m
Answer» Given: A BODY is released from the TOP of the tower of HEIGHT 100m . After 2 seconds it is stopped and then instantaneously released. To find: Height of body after next 3 seconds. Calculation: After 2 seconds , the height of the BALL above the earth will be: $\therefore \: h = 100 - y$ $= > \: h = 100 - \bigg \{ut + \dfrac{1}{2} g {t}^{2} \bigg \}$ $= > \: h = 100 - \bigg \{0 + \dfrac{1}{2} g {t}^{2} \bigg \}$ $= > \: h = 100 - \bigg \{ \dfrac{1}{2} g {t}^{2} \bigg \}$ $= > \: h = 100 - \bigg \{ \dfrac{1}{2} g {(2)}^{2} \bigg \}$ $= > h = 100 - 20$ $= > h = 80 \: m$ Its stopped for an INSTANT and again released. Let height after next 3 seconds be H. $\therefore \: H = 80 -( y2)$ $= > \: H = 80 - \bigg \{u(t2) + \dfrac{1}{2} g {(t2)}^{2} \bigg \}$ $= > \: H = 80 - \bigg \{0 + \dfrac{1}{2} g {(t2)}^{2} \bigg \}$ $= > \: H = 80 - \bigg \{ \dfrac{1}{2} g {(t2)}^{2} \bigg \}$ $= > \: H = 80 - \bigg \{ \dfrac{1}{2} g {(3)}^{2} \bigg \}$ $= > \: H = 80 - \bigg \{ 45\bigg \}$ $= > \: H = 35 \: m$ So, final answer is : $\boxed{ \red{ \bold{ \large{ \: H = 35 \: m}}}}$