An aeroplane is flying along the horizonyal at a velocity Vo starts to ascend describing a circle with height ‘h’ above the initial level of motion acc. To the expression v2 =vo 2-2ßh the velocity of plane at the upper point of trajectory is v1=V0/2. Find the acceleration magnitude of the plane when its velocity is directed vertically upwards.

Question

Sandeep Pathak · Answer

The velocity of the aeroplane is vertical when it has completed quarter circle. In that case, there are two components of acceleration:

Centripetal acceleration, a_cdirected towards center, i.e. horizontal.
Tangential acceleration, a_t= g, directed downwards.

Let’s assume the velocity be v₁when the aeroplane is moving vertically. Then,
$v_1^2 = v_0^2-2g\frac{h}{2} = v_0^2-gh$
Also, it is given
$\left(\frac{v_0}{2}\right)^2 = v_0^2-2gh$
This gives
$v_1^2 = \frac{5gh}{3}\\ a_c=\frac{v_1^2}{h/2}=\frac{10g}{3} \Rightarrow a = \sqrt{a_c^2+a_t^2}=\frac{\sqrt{109} g}{3}$

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