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2.Motion in Straight Line
hard
એક બલૂન $4.9 m/s^2$ ના પ્રવેગથી ઉપર તરફ ગતિ કરવાનું શરૂ કરે છે. $2 sec$ પછી તેમાંથી પથ્થર મુકત કરતાં પથ્થરે પ્રાપ્ત કરેલી મહત્તમ ઊંચાઇ?.........$m$ $(g = 9.8\,m/{\sec ^2})$
A
$14.7$
B
$19.6$
C
$9.8$
D
$24.5 $
Solution
(a) Height travelled by ball (with balloon) in $2$ sec
${h_1} = \frac{1}{2}a\;{t^2} = \frac{1}{2} \times 4.9 \times {2^2} = 9.8\;m$
Velocity of the balloon after 2 sec
$v = a\;t = 4.9 \times 2 = 9.8\;m/s$
Now if the ball is released from the balloon then it acquire same velocity in upward direction.
Let it move up to maximum height ${h_2}$
${v^2} = {u^2} – 2g{h_2}$ $⇒$ $0 = {(9.8)^2} – 2 \times (9.8) \times {h_2}$$\therefore $${h_2}=4.9\,m$
Greatest height above the ground reached by the ball $ = {h_1} + {h_2} = 9.8 + 4.9 = 14.7\;m$
Standard 11
Physics
