Question

# A particle is projected from ground level with an initial velocity of 35m/s at an angle of tan−13/4 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)t−12gt2 2=21t−5t2 5t2−21t+2=0 The time can be written as, t1+t2=215 t1t2=25 (t1−t2)2=(t1+t2)2−4t1t2 (t1−t2)2=(215)2−4×25 (t1−t2)2=(215)2−4×25 (t1−t2)=4sec Thus, the time for which the particle is more than 2m above the ground is 4sec.

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