Question

# A snowball rolls off a barn roof that slopes downward at an angle of α= 45.0∘...

A snowball rolls off a barn roof that slopes downward at an angle of α= 45.0∘ . (See the figure below) The edge of the roof is H= 17.0 m above the ground, and the snowball has a speed of v = 4.00 m/s

as it rolls off the roof. Ignore air resistance.

How far from the edge of the barn does the snowball strike the ground if it doesn't strike anything else while falling?

Solution-

Using the formula

Y = VyT + (1/2)gT^2

here

Y is the vertical distance that snowball has to fall = 17m
Vy is the vertical component of the initial velocity = 4.00(sin 45)

= 2.83 m/sec.
T is the time for the snowball to hit the ground
g (acceleration due to gravity) = 9.8 m/sec^2

Substituting the values,

17 = 2.83T + (1/2)(9.8)T^2

17 = 2.83 T + 4.9T^2

4.9T^2 + 2.83T - 17 = 0

T = 1.596 sec.

Now using formula
X = Vx(T)

here

X is the distance from the edge of the barn where the snowball will land
Vx is the horizontal component of the initial velocity = 4(cos 45)

= 2.83 m/sec.
T = 1.596 sec
Substituting values,

X = 2.83 * 1.596

X = 4.52 meters