First slide

Projection Under uniform Acceleration

Question

Two particles of same mass are projected simultaneously with same speed 20 ms^–1 from the top of a tower of height 20m. One is projected vertically upwards and other projected horizontally. The maximum height attained by centre of mass from the ground will be (g = 10 ms^–2)

Moderate

A

10√2m
B

25m
C

25√2m
D

5m

Solution

$\large {{\bar v}_{cm}} = \frac{{{m_1}{{\bar v}_1} + {m_2}{{\bar v}_2}}}{{{m_1} + {m_2}}}$
$\large = \frac{{m\left( {20\hat j} \right) + m\left( {20\hat i} \right)}}{{2m}}$
$\large {v_{cm}} = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2}} = 10\sqrt 2 m/s$
$\large Tan\theta = \frac{{{v_1}}}{{{v_2}}} = 1\, \Rightarrow \theta = {45^0}$
$\large {H_{\max }} = \frac{{v_{cm}^2{{\sin }^2}\theta }}{{2g}} = \frac{{{{\left( {10\sqrt 2 } \right)}^2} \times {{\sin }^2}\left( {{{45}^0}} \right)}}{{2 \times 10}} = 5m$
from ground COM distance = h + H_max
= 20 + 5 = 25m

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