First slide

Mechanical energy conservation

Question

A body projected obliquely with velocity 19.6 m/s has its kinetic energy at the maximum height equal to 3 times its potential energy there. Its position after 1second of projection from the ground is (h = maximum height)

Moderate

A

h/2
B

h/4
C

h/3
D

h

Solution

$K E_{a t m a x i m u m h e i g h t} = 3 P E_{a t m a x i m u m h e i g h t}$
$\frac{1}{2} m u^{2} C o s^{2} θ = 3 m g [\frac{u^{2} S i n^{2} θ}{2 g}]$
$C o s^{2} θ = 3 S i n^{2} θ$ hence $θ = 30^{0}$
$y = u S i n θ t - \frac{1}{2} g t^{2}$
y= $19.6 \frac{1}{2} \times 1 - \frac{1}{2} 9.8 {(1)}^{2}$ =4.9 m
$h = \frac{u^{2} S i n^{2} θ}{2 g} = \frac{{(19.6)}^{2}}{2 \times 9.8} \times \frac{1}{4} = 4.9 m$
so y=h

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