Question

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

Moderate

Solution

Consider the diagram
For horizontal motion, 50 = 10 cos 30° x t
$\Rightarrow t = \frac{5}{\cos 30 °} = \frac{5}{\frac{\sqrt{3}}{2}} = \frac{10}{\sqrt{3}}$
For vertical motion,
$h = 10 \sin 30 ° \times \frac{10}{\sqrt{3}} - \frac{1}{2} \times 10 \times {(\frac{10}{\sqrt{3}})}^{2}$
$= 10 \times \frac{1}{2} \times \frac{10}{\sqrt{3}} - \frac{100}{\sqrt{3}} \times \frac{5}{\sqrt{3}} = \frac{100}{\sqrt{3}} (\frac{1}{2} - \frac{5}{\sqrt{3}})$
$= \frac{50}{\sqrt{3}} [1 - \frac{10}{\sqrt{3}}] m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME