Question

An object is thrown towards the tower which is at a horizontal distance of $50m$ with initial velocity of $10{ms}^{-1}$ and making an angle ${30}^{\circ }$ with the horizontal. The object hits the tower at certain height. The height from the bottom of the tower where the object hit the tower is ($g=10{ms}^{-2}$) :

• A. $\frac{50}{\sqrt{3}}[1-\frac{10}{\sqrt{3}}]m$
• B. $\frac{50}{3}[1-\frac{10}{\sqrt{3}}]m$
• C. $\frac{100}{\sqrt{3}}[1-\frac{10}{\sqrt{3}}]m$
• D. $\frac{100}{3}[1-\frac{10}{\sqrt{3}}]m$

Answer 1

The correct option is A $\frac{50}{\sqrt{3}}[1-\frac{10}{\sqrt{3}}]m$

Time in which the body covers the horizontal distance of $50m=\frac{50}{v_x}=\frac{50}{vcos\theta }=\frac{50}{10\times cos{30}^{\circ }}$

$=\frac{10}{\sqrt{3}}s$

Thus, the height that the object reaches in this time = $s=v_{y}t-\frac{1}{2}g{t}^2$

$=vsin\theta t-\frac{1}{2}g{t}^2$
$=10\times \frac{1}{2}\times \frac{10}{\sqrt{3}}-\frac{1}{2}\times 10\times (\frac{10}{\sqrt{3}}{)}^2$

$=\frac{50}{\sqrt{3}}-\frac{500}{3}$

$=\frac{50}{\sqrt{3}}[1-\frac{10}{\sqrt{3}}]m$