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Question

A particle is projected from ground level with an initial velocity of 35m/s at an angle of tan13/4 to the horizontal. Find the time for which the particle is more than 2m above the ground. [g=10m/s2]

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Solution

The angle is given as,

tanθ=34

θ=36.86

The height is given as,

h=(usinθ)t12gt2

2=21t5t2

5t221t+2=0

The time can be written as,

t1+t2=215

t1t2=25

(t1t2)2=(t1+t2)24t1t2

(t1t2)2=(215)24×25

(t1t2)2=(215)24×25

(t1t2)=4sec

Thus, the time for which the particle is more than 2m above the ground is 4sec.


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