Question

From the top of a tower 49 m high a ball is thrown with a velocity of 30 m/sec at an elevation of 60° with the horizontal .When will it strike the ground?​

Answer 1

Answer:

• It will strike at t = 6.55 seconds

Given:

1. Height of tower (H) = 49 m.
2. Initial velocity (u) = 30 m/s.
3. Angle of projection (θ) = 60°

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\bigstar\;\underline{\boxed{\sf H=\bigg[-u\;\sin\theta\bigg]\; t+\dfrac{1}{2}\;gt^{2}}}$

Here in this case we need to take positive and negative Cartesian system. So, let's consider the downward direction be Positive,and upward to be negative. therefore acceleration due to gravity will be positive as it acts downward and vertical component will be negative as it will be on the upward direction.

Substituting the values,

$\longrightarrow\sf 49=\bigg[-30\times \sin 30^{\circ}\bigg]\;t+\dfrac{1}{2}\times10\times t^{2}\\\\\\\longrightarrow\sf 49=\Bigg[-30\times\dfrac{\sqrt{3}}{2}\Bigg]\;t+5\;t^{2}\\\\\\\longrightarrow\sf 49=-15\sqrt{3}\times t+5\;t^{2}\\\\\\\longrightarrow\sf 49=-15\sqrt{3}\;t+5\;t^{2}\\\\\\\longrightarrow\sf 5\;t^{2}-15\sqrt{3}\;t-49=0\\\\\\\longrightarrow\boxed{\sf 5\;t^{2}-15\sqrt{3}\;t-49=0}$

Hence we got a quadratic equation!

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Solving it by quadratic formula,

$\bigstar\;\underline{\boxed{\sf t=\dfrac{-b\;\pm\;\sqrt{b^{2}-4\;ac}}{2\;a}}}$

Substituting and solving,

$\longrightarrow\sf t=\dfrac{-\bigg(-15\sqrt{3}\bigg)\;\pm\;\sqrt{\bigg(15\sqrt{3}\bigg)^{2}-\bigg(4\times5\times-49\bigg)}}{2\times 5}\\\\\\\\\longrightarrow\sf t=\dfrac{15\sqrt{3}\;\pm\;\sqrt{\bigg(225\times 3\bigg)-\bigg(-980\bigg)}}{10}\\\\\\\\\longrightarrow\sf t=\dfrac{15\sqrt{3}\;\pm\;\sqrt{\bigg(675+980\bigg)}}{10}\\\\\\\\\longrightarrow\sf t=\dfrac{15\sqrt{3}\;\pm\;\sqrt{\bigg(1665\bigg)}}{10}$

Here we got two equations,

$\begin{array}{c c c}&\sf{Case\colon 1}&\sf{Case\colon 2}\\\\\longrightarrow&\sf{t=\dfrac{15\sqrt{3}\;+\;\sqrt{\bigg(1665\bigg)}}{10}}&\sf{t=\dfrac{15\sqrt{3}\;-\;\sqrt{\bigg(1665\bigg)}}{10}}\\\\\\\\\longrightarrow&\sf{t=\dfrac{\bigg(15\times 1.732\bigg)+40.6}{10}}&\sf{t=\dfrac{\bigg(15\times 1.732\bigg)-40.6}{10}}\\\\\\\\\longrightarrow&\sf{t=\dfrac{\bigg(25.9\bigg)+40.6}{10}}&\sf{t=\dfrac{\bigg(25.9\bigg)-40.6}{10}}\\\\\\\\\longrightarrow&\sf{t=\dfrac{66.5}{10}}&\sf{t=\dfrac{-14.7}{10}}\end{array}$

Time cannot be negative, so

$\longrightarrow\sf t=\dfrac{65.5}{10}\\\\\\\longrightarrow\sf t=6.55\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf t=6.55\;sec}}}}$

∴ It will strike ground at t = 6.55 seconds .

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Alternate Way:

In these cases we have a formula for time period.

$\bigstar\;\underline{\boxed{\sf T=\dfrac{u\;\sin\theta}{g}\;\pm\;\sqrt{\dfrac{u^{2}\;\sin^{2}\theta}{g^{2}}+\dfrac{2\;h}{g}}}}$

Substituting the values,

$\longrightarrow\sf T=\dfrac{30\times \sqrt{3}}{10\times 2}\;\pm\;\sqrt{\dfrac{(30)^{2}}{(10)^{2}}\times\dfrac{3}{4}+\dfrac{2\times 49}{10}}\\\\\\\\\longrightarrow\sf T=\dfrac{3\; \sqrt{3}}{2}\;\pm\;\sqrt{\dfrac{900}{100}\times\dfrac{3}{4}+\dfrac{98}{10}}\\\\\\\\\longrightarrow\sf T=\dfrac{3\; \sqrt{3}}{2}\;\pm\;\sqrt{9\times\dfrac{3}{4}+\dfrac{98}{10}}\\\\\\\\\longrightarrow\sf T=\dfrac{3\; \sqrt{3}}{2}\;\pm\;\sqrt{\dfrac{27}{4}+\dfrac{98}{10}}$

$\longrightarrow\sf T=\dfrac{3\; \sqrt{3}}{2}\;\pm\;\sqrt{\dfrac{135+196}{20}}\\\\\\\longrightarrow\sf T=\dfrac{3\; \sqrt{3}}{2}\;\pm\;\sqrt{\dfrac{331}{20}}\\\\\\\longrightarrow\sf T=\dfrac{3\; \sqrt{3}}{2}\;\pm\;\sqrt{\dfrac{331}{4\times 5}}\\\\\\\longrightarrow\sf T=\dfrac{3\; \sqrt{3}}{2}\;\pm\;\dfrac{1}{2}\sqrt{\dfrac{331}{5}}\\\\\\\longrightarrow\sf T=\dfrac{3\; \sqrt{3}}{2}\;\pm\;\dfrac{\sqrt{66.2}}{2}$

Solving,

$\begin{array}{c c c}&\sf{Case\colon 1}&\sf{Case\colon 2}\\\\\longrightarrow&\sf{t=\dfrac{3\sqrt{3}\;+\;\sqrt{\bigg(66.2\bigg)}}{2}}&\sf{t=\dfrac{3\sqrt{3}\;-\;\sqrt{\bigg(66.2\bigg)}}{2}}\\\\\\\\\longrightarrow&\sf{t=\dfrac{\bigg(3\times 1.732\bigg)+8.13}{2}}&\sf{t=\dfrac{\bigg(3\times 1.732\bigg)-8.13}{2}}\\\\\\\\\longrightarrow&\sf{t=\dfrac{\bigg(5.19\bigg)+8.13}{2}}&\sf{t=\dfrac{\bigg(5.19\bigg)-8.13}{2}}\\\\\\\\\longrightarrow&\sf{t=\dfrac{13.31}{2}}&\sf{t=\dfrac{-2.94}{2}}\end{array}$

Time cannot be negative,

$\longrightarrow\sf t=\dfrac{13.31}{2}\\\\\\\longrightarrow\sf t=6.55\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf t=6.55\;sec}}}}$

∴ It will strike ground at t = 6.55 seconds .

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$\rule{300}{1.5}$

Answer 2

Answer

It will strike at t = 6.55 seconds.