# Physics

posted by on .

3. A projectile is launched at 10.0 m/s in a direction 45.0 degrees above the horizon. It hits the ground 10.0 m below its initial position. What is its speed just before hitting the ground? Ignore air resistance.

The answer is supposed to be 17.2 m/s.

• Physics - ,

I've been using:

v(f)^2 = v(i)sin(45)^2 - 2g(y-y(i))

Then I use:

v = sqrt((v(f)^2 + v(i)^2)

However, doing this gives me 18.6 m/s instead of 17.2 m/s. Any help would be greatly appreciated.

• Physics - ,

I've found that:

v(f)^2 = v(i)^2 -2g(y - y(i))

gives me 17.2 m/s. However, this way doesn't account for the angle, so I don't think it's the right way to solve this.

• Physics - ,

You've to consider the motion in vertical and horizontal directions separately.
A)Vertical (y) dir.=>
Vy^2 = Uy^2 + 2*(-g)*s
= (10Sin45)^2 + 2*(-9.8)(-10)
= 50 + 196 = 246
So, Vy= 15.68 m/s
(Vy is the vertical component of the projectile when it hits the ground)
B) Horizontal (x)dir. =>
Vx= Ux = 10Cos45 = 7.07 m/s

Projectile's vel. just before it hits the gnd.=>
V = sqrt(Vy^2 + Vx^2)
= sqrt(246 + 50)
= 17.2 m/s