Question 1040257
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A body is projected downward at an angle of 30 to the horizontal with a velocity of 9.8 m/s from the top 
of the tower of 29.4 m high. How long will it take before striking the ground?
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<pre>
The body moves uniformly at the constant speed in horizontal direction and at the constant acceleration 
g = {{{9.8}}} {{{m/s^2}}} downward in vertical direction.
     (! Do not miss the initial velocity 9.8 {{{m/s}}} with the gravity acceleration g = {{{9.8}}} {{{m/s^2}}} !)

To answer the question, it is enough to consider vertical movement of the body.

Its initial vertical velocity is  v = {{{sin(30^o)*9.8}}} = {{{(1/2)*9.8}}} = {{{4.9}}} {{{m/s}}} directed downward.

Then an equation for the vertical coordinate of the body h(t) is

h(t) = {{{-g/2*t^2 + v*t + 29.4}}},

where g = {{{9.8}}} {{{m/s^2}}}  and  v = {{{-4.9}}} {{{m/s}}},  or

h(t) = {{{-4.9*t^2 - 4.9*t + 29.4}}}.

Here h(t) is the vertical distance from the body to the ground level in meters, t is time in seconds.

To find the time moment when the body strikes the ground, you need to solve a quadratic equation 

h(t) = 0,  or  {{{-4.9*t^2 - 4.9*t + 29.4}}} = {{{0}}}.

It is the same as 

{{{4.9*(t^2 + t - 6)}}} = {{{0}}}. 

The last equation is equivalent to 

{{{t^2 + t - 6}}} = {{{0}}}.

Factor it. It is equivalent to

(t-3)*(t+2) = 0

and has the roots t = 3 and t = -2.

Only positive root is acceptable.

So, the answer is: t = 3.  The body hits the ground in 3 seconds.
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