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A body is projected up such that its position vector varies with time as $r = \{3 t \hat{i} + (4 t - 5 t^{2}) \hat{j}\} m$ . Here, t is in seconds. Find the time and x-coordinate of particle when its y-coordinate is zero.

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at t=0, x=0

And at t=0.8 s, x=2.4 m

at t=0, x=10

And at t=0.8 s, x=2.4 m

at t=0, x=0

And at t=0.8 s, x=20 m

at t=0, x=10

And at t=0.8 s, x=0 m

answer is A.

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Detailed Solution

y-coordinate of particle is zero when

$4 t - 5 t^{2} = 0$

Solving, $t = 0 s and t = 0.8 s$

Now, x = 3t (given)

When t = 0 s, x = 0 m,

and when t = 0.8 s, x = 2.4 m

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