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# A stone is thrown up with an initial velocity V● , there is resisting acceleration - Kv due to air where v is instantaneous velocity and K is some positive constant find First time taken to reach the highest point .Second maximum height attained

2 years ago

Susmita
425 Points

When it reaches the highest point it stops for a moment and then starts falling off.So at maximum height speed is zero.
Equation of force is
mf=-(mg+kv)
Or,mdv/dt=-(mg+kv)
Or,mdv/(mg+kv)=-dt
Integrate dv from v to 0 and dt from 0 to t.
(m/k)log(mg/(mg+kv))=-t
Or,t=(m/k)log((mg+kv)/mg)
This the value of time.
Now write v from the expression.You will get
$\frac{kv}{mg}=e^\frac{kt}{m}-1$
Now write v as dx/dt.
$\frac{kdx}{mg}=(e^\frac{kt}{m}-1)dt$
Integrate dx from 0 to h and dt from 0 to t.
$\frac{kh}{mg}=\frac{m}{k}(e^\frac{kt}{m}-1)-t$
Put value of t and exp(kt/m) as obtained earlier to obtain h.

2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions