Question

A projectile is thrown from ground level with an initial velocity 4i + 3j. It reaches its greatest height above
ground level after-
(C) 0.4 s
(D) 3 s
(A) 0.2 s
(B) 0.3 s​

Answer 1

Answer:

0.3 seconds

Explanation:

upward velocity = 3j which is 3 m/s

upward direction is given as usinθ = 3 m/s

$T = \frac {2usin}{g}$

$T = \frac{2*3}{10}$

$T = \frac{6}{10}$

T = 0.6 sec      ------------ (Total time of the travel)

we know that the time taken for body to move from initial to final point is 0.6 seconds also,

the maximum height of the body is always attained at range/2

i.e. to prove it first draw a projectile and draw perpendicular from it's maximum height to a point intersecting it's range

the intersection is always exactly present between the initial and final point of the projectile so we conclude that max height position is always range/2

therefore total time taken to travel Range = 0.6 sec

so,  

total time taken to travel Range/2 is ,

0.6/2

which is

0.3 seconds

HOPE IT HELPS

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