Question

A body is projected vertically up at t=0 with a velocity of 98 m/s .another body is projected from the same point with same velocity after 4 seconds.both bodies will meet at t= ?

Answer 1

Answer:

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The body thrown upward. Velocity of body = 98 m/s

Initial velocity = 0, final velocity = 4 secs

Velocity of the body after 4 secs, v = u + at

a = -9.8$m/s^2$

v = 98 - (9.8)(4) = 98 - 39.2 = 58.8 m/s

height of the body after 4 secs h = $ut-\frac { 1 }{ 2 } g{ t }^{ 2 }$

$h\quad =\quad 98(4)-\frac { 1 }{ 2 } (9.8){ (4) }^{ 2 }$

h = 392 - 78.4 = 313.6 m

As the direction of the body is in the same direction, relative velocity =98 - 58.8 = 39.2 m/s

Time = height / relative velocity

Time = 313.6/39.2 = 8 seconds.