Posted by **Nikitha** on Saturday, September 1, 2012 at 8:04am.

Q Exactly 2.7s after a projectile is fired into the air from the ground ,it is observed to have a velocity v = (8.8i+4.5 j) m/s where the x-axis is horizontal and the y axis is positive upward .

a)Determine the horizontal range of the projectile

b)Determine the maximum height above the ground

c)Determine the speed of motion just above the projectile strikes the ground .

d)Determine the angle of the motion just before the projectile strikes the ground .

- Physics -
**Adam**, Saturday, September 15, 2012 at 11:01pm
a) The horizontal component of any velocity remains constant as it goes through the air, so the horizontal component of the initial velocity would still be 8.8i. The distance can be calculated using the formula Xf=Vx*t:

Xf=8.8*2.7=23.76

b)Maximum height is given by the formula:

H_max=((Vo*sin(a)))^2/2g

The initial velocity can be found by finding the initial y component using the formula Vf=Vo+at

4.5=Vy-9.81*2.7=>Vy=30.987

Now Vo=sqrt(Vx^2+Vy^2)=sqrt(8.8^2+30.987^2)

Now you can find the angle between the horizontal and the firing velocity and plug it in the to formula given above.

c) and d) after the projectile as travelled the entire trajectory back to the ground, the speed and the angle the projectile has is equal to the initial velocity and angle

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