A ball is thrown upwards with a speed of $$40$$ $$m/s$$. When the speed becomes half of the initial speed, gravity is switched off for next $$2$$ second. After that gravity is again switched on but magnitude of gravity is doubled. The total distance travelled by the ball from t $$=$$ $$0$$ to the time when the ball reaches the maximum height (in$$ $$m) isTake $$g=10$$ $$ms-2$$

Question

A ball is thrown upwards with a speed of $$40$$ $$m/s$$. When the speed becomes half of the initial speed, gravity is switched off for next $$2$$ second. After that gravity is again switched on but magnitude of gravity is doubled. The total distance travelled by the ball from t $$=$$ $$0$$ to the time when the ball reaches the maximum height (in$$ $$m) isTake $$g=10$$ $$ms-2$$

Infinity Learn · Accepted Answer

Speed will remain half or $$20$$ msafter $$2$$ s.$$d1=ut+12at2=40$$×$$2-12$$×$$10$$×$$(2)2=60$$ $$md2=vt=20$$×$$2=40$$ $$md3=v22g$$'$$=(20)22$$×$$20=10$$ m ∴$$d=d1+d2+d3=110$$ m

Rectilinear Motion

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Speed will remain half or 20 $\frac{m}{s}$ after 2 s.
$d_{1} = ut + \frac{1}{2} {at}^{2} = 40 \times 2 - \frac{1}{2} \times 10 \times {(2)}^{2} = 60 m$
$d_{2} = vt = 20 \times 2 = 40 m$
$d_{3} = \frac{v^{2}}{2 g'} = \frac{{(20)}^{2}}{2 \times 20} = 10 m$
$∴ d = d_{1} + d_{2} + d_{3} = 110 m$

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Rectilinear Motion

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Speed will remain half or 20 msafter 2 s.d1=ut+12at2=40×2-12×10×(2)2=60 md2=vt=20×2=40 md3=v22g'=(20)22×20=10 m ∴d=d1+d2+d3=110 m

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Speed will remain half or 20 $\frac{m}{s}$ after 2 s.
$d_{1} = ut + \frac{1}{2} {at}^{2} = 40 \times 2 - \frac{1}{2} \times 10 \times {(2)}^{2} = 60 m$
$d_{2} = vt = 20 \times 2 = 40 m$
$d_{3} = \frac{v^{2}}{2 g'} = \frac{{(20)}^{2}}{2 \times 20} = 10 m$
$∴ d = d_{1} + d_{2} + d_{3} = 110 m$