Solve this-A body is thrown vertically upwards- If constant air resistance is to be taken into account- then the time nbsp-during which the body rises is t1 and the time of fall nbsp-is t2 then nbsp-1- nbsp-t1-t2-2- nbsp- nbsp-t1 lt- nbsp-t2-3- nbsp- nbsp-t1 nbsp- gt- -t2-4- nbsp- nbsp-t1-2t2

Dear Student , Let #160;F #160;be #160;the #160;force #160;of #160;air #160;resistance #160;.For #160;upward #160;motion #160;F #160;act ...

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Byju's Answer

Standard IX

Physics

Equation of Motion of Free Fall

Solve this: ...

Question

Solve this:
A body is thrown vertically upwards. If constant air resistance is to be taken into account, then the time
during which the body rises is t₁ and the time of fall is t₂ then
(1) t₁=t₂

(2) t_1<t₂

(3) t₁> t₂
(4) t₁=2t₂

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Solution

Dear Student ,
$L e t F b e t h e f o r c e o f a i r r e s i s \tan c e . F o r u p w a r d m o t i o n F a c t s d o w n w a r d s . L e t a_{1} = R e t a r d a t i o n . m g + F = m a_{1} F o r t h e d o w n w a r d s m o t i o n F a c t s u p w a r d s . L e t a_{2} b e t h e d o w n w a r d s a c c e l e r a t i o n . m g - F = m a_{2} \Rightarrow a_{2} = g - \frac{F}{m} T h u s, a_{1} > a_{2} T h e r e f o r e, t_{1} < t_{2} B e c a u s e, h = \frac{1}{2} a_{1} {t_{1}}^{2} a n d h = \frac{1}{2} a_{2} {t_{2}}^{2}$

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Solve this: A body is thrown vertically upwards. If constant air resistance is to be taken into account, then the time during which the body rises is t1 and the time of fall is t2 then (1) ​t1=​t2 (2) ​ ​t1< t2 (3) ​ ​t1 > ​t2 (4) ​ ​t1=​2t2

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Solve this:
A body is thrown vertically upwards. If constant air resistance is to be taken into account, then the time
during which the body rises is t₁ and the time of fall is t₂ then
(1) t₁=t₂

(2) t_1<t₂

(3) t₁> t₂
(4) t₁=2t₂