# CBSE Class 11-science Answered

**a body is projected with speed 40root2 m/s downwards an angle of 45 degree from vertical if h=100 m find 1. time of flight 2. range**

Asked by sohail.saba0311 | 09 Sep, 2021, 08:23: PM

Expert Answer

As shown in above figure, an object is projected downward from a height h = 100 m above ground level

with velocity u 40√2 m/s so that angle of projection is 45

^{o}with vertical.Projection velocity is resolved into two components,

(i) horizontal component u

_{x}= 40 m/s and (ii) vertical component u_{y}= 40 m/sTime of flight t , i.e., time taken for the object to reach ground is determined from the following equation

h = u

_{y}t + [ (1/2) g t^{2}] .................... (1)where g is acceleration due to gravity .

By substituting values in eqn.(1) , we get the following quadratic equation

100 = 40 t + 4.9 t

^{2}or 4.9 t^{2}+ 40 t -100 = 0By solving above quadratic equation, we get t = 2 s

Range R , i.e. Horizontal distance from projection point is determined by following equation

R = u

_{x}twhere t is time of flight

Hence, we get , R = 40 × 2 = 80 m

Answered by Thiyagarajan K | 09 Sep, 2021, 09:59: PM

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