Question

A particle is projected with a velocity 19.6 metre per second making an angle 30 degree with the horizontal it reaches the highest point in which time ​

Answer 1

Answer:

1 second

Explanation:

In oblique projectile motion

Time taken to reach maximum height is

T = uSin∅ / g

= (19.6 m/s × Sin30) / (9.8 m/s²)

= (9.8 m/s) / (9.8 m/s²)

= 1 second

Answer 2

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A particle is projected with a velocity 19.6 metre per second making an angle 30 degree with the horizontal it reaches the highest point in which time ??

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$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \bf{\boxed{\huge\mathcal{1 second}}}$

______________________________________________

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$t = \frac{u \: sin(x)}{g}$

where..

t=time to reach the maximum height

X=angle of projection

g=9.8 m/sec²

$\:\: \underline{\underline{\bf{\large\mathfrak{~~solution~~}}}}$

therefore the time is

$= \frac{19.6 \: sin(30)}{9.8} \\ = \frac{9.8}{9.8} \\ = 1 \: sec$

Hope this helps you