A man stands on the roof of a 15.4 m tall building and throws a rock with a speed of 29.0 m/s at an angle of 33.3 degrees above the horizontal. Calculate

Question

A) the maximum height above the roof that the rock reaches
B) the speed of the rock just before it strikes the ground; and
C) the horizontal range from the base of the building to the point where the rock strikes the ground.

Answer 1

free fall equation ... h = -1/2 g t^2 + [29.0 sin(33.3º) t] + 15.4 m

the max height occurs at the vertex of the parabola ... on the axis of symmetry
... correct for the height of the building

the horizontal speed is constant ... 29.0 cos(33.3º) m/s

the vertical speed at impact is ... √(2 g 15.4) m/s downward

the two speed components add as vectors ... a^2 + b^2 = c^2

the range is the horizontal speed divided by the time of flight
... plug in zero for h in the free fall equation to find t

Answer 2

Vo = 29m/s[33.3o].
Xo = 29*Cos33.3 = 24.2 m/s.
Yo = 29*sin33.3 = 15.9 m/s.

A. Y^2 = Yo^2 + 2g*h = 0,
15.9^2 + (-19.6)h = 0,
h = 12.9 m.

B. Y^2 = Yo^2 + 2g*h = 0 + 19.6 * (12.9+15.4) = 554.7,
Y = 23.6 m/s.
V = sqrt(Xo^2 + Y^2) = sqrt(24.2^2 + 23.6^2) =

C. Y = Yo + g*Tr = 0,
15.9 + (-9.8)Tr = 0,
Tr = 1.62 s. = Rise time.

0.5g*Tf^2 = (15.4 + 12.9),
Tf = 2.40 s. = Fall time.

Range = Xo(Tr + Tf)

Range = sqrt(Xo^2 + Y^2) = sqrt(24.2^2 + 23.6^2) =