A body is projected downwards at an angle of $30^{o}$ to the horizontal with a velocity of $9.8 m / s$ from the top of a tower $29.4 m$ high. How long will it take before striking the ground?

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Solution

Initial vertical velocity $u_{x} = 9.8 s i n 30^{0} = 4.9 m / s$
put this value in $h = u_{x} t + g t^{2} / 2$ use $h = 29.4 m$ and $g = 9.8 m / s^{2}$
$29.4 = 4.9 t + g t^{2} / 2$
we get $t = 3$ $s e c o n d$

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A body is projected downwards at an angle of 30o to the horizontal with a velocity of 9.8m/s from the top of a tower 29.4m high. How long will it take before striking the ground?

Initial vertical velocity ux=9.8sin300=4.9m/sput this value in h=uxt+gt2/2 use h=29.4m and g=9.8m/s229.4=4.9t+gt2/2we get t=3 second

A body is projected downwards at an angle of $30^{o}$ to the horizontal with a velocity of $9.8 m / s$ from the top of a tower $29.4 m$ high. How long will it take before striking the ground?

Initial vertical velocity $u_{x} = 9.8 s i n 30^{0} = 4.9 m / s$
put this value in $h = u_{x} t + g t^{2} / 2$ use $h = 29.4 m$ and $g = 9.8 m / s^{2}$
$29.4 = 4.9 t + g t^{2} / 2$
we get $t = 3$ $s e c o n d$