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Projectile Motion Q

 
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prosteve037



Joined: 25 Sep 2010
Posts: 6
Location: 222

PostPosted: Sat Sep 25, 2010 7:27 pm    Post subject: Projectile Motion Q Reply with quote

Q - A brick is thrown upward from the top of a building at an angle of 20 to the horizontal and with an initial speed of 12 m/s. If the brick is in flight for 2.7 s, how tall is the building?

A - So...

Vi = 12 m/s
t = 2.7 s
= 20

Right? But now what? lol

Please help. I don't know how to solve this problem at all Crying or Very sad
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hepcj
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Joined: 23 Jun 2007
Posts: 125

PostPosted: Sat Sep 25, 2010 8:01 pm    Post subject: Reply with quote

With this problem you know the initial velocity of the brick but it is not vertically upward. From your class you should have learnt that the horizontal motion and vertical motion are independent so we need to know what is the vertical component of the velocity and this is where angle comes in.

u = 12 cos(20 deg)

We know the total time of flight of the brick is 2.7 secs and this must comprise the time it take to reach maximum height and fall back down to the ground.

From the equations of linear motion we can find the distance.
s = u t + 0.5 g t^2 (equation 1) travelled by inserting the time (2.7s).

where s is the distance, u is the initial velocity, g is the acceleration due to gravity (remember, it acts in the opposite direction to the velocity), and t is the time.

The absolute magnitude of this should be the height of the building.
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