University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises - Page 98: 3.74

Answer

(a) Bolt will hit the floor after $t=0.78 \space s$. (b) Speed of the bolt w.r.t. observer in the lift, $v_{bl} =7.64 \space m/s$. (c) Speed of the bolt w.r.t. observer on the ground, $v_b=5.14 \space m/s$. (d) Displacement of the bolt w.r.t. observer on the ground, $s_b=1.03 \space m$ (downward direction).

Work Step by Step

Motion of the bolt relative to lift (assuming downward direction to be positive), Initial relative velocity $u_{bl}=0$, relative acceleration $a_{bl}=a_{b}-a_{l}=g-0=g$, relative displacement $s_{bl}=3 \space m$. (a) Using equation $s=ut+\frac{1}{2}at^2$ for bolt w.r.t. lift, $-3=0\times t+\frac{1}{2}(9.8)t\Rightarrow t=\sqrt\frac{2\times 3}{9.8}=0.78 \space s$ (b) Speed of the bolt w.r.t. observer in the lift, $v_{bl}=u_{bl}+a_{bl}t\Rightarrow v_{bl}=0+9.8\times 0.78=7.64 \space m/s$ (c) Speed of the bolt w.r.t. observer on the ground, $v_b=u_b+a_bt\Rightarrow v_{bl}=-2.50+9.8\times 0.78=5.14 \space m/s$ (d) Displacement of the bolt w.r.t. observer on the ground, $s_b=u_bt+\frac{1}{2}a_bt^2\Rightarrow s_b=-2.50\times 0.78+\frac{1}{2}(9.8)(0.78)^2=1.03 \space m$ (downward direction)
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