A 0.56 kg rock is projected from the edge of the top of a building with an initial velocity of?

Question

A 0.56 kg rock is projected from the edge of
the top of a building with an initial velocity of
10.2 ms at an angle 44degrees above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 17.8 m from the base
of the building.

Assume: The ground is level and that the
side of the building is vertical. The accelera-
tion of gravity is 9.8 ms^2 :
How tall, h, is the building? Answer in
units of m.

Answer 1

IGNORE WEIGHT
the horizontal velocity is constant at 10.2cos44 = 7.337
so 7.337t = 17.8, t =2.426s.

for the vertical velocity, it is 10.2sin44 = 7.086
the equation for height is
h = y0 + v0t +-1/2 g t2
set origin at the ground....
0 = y0 + 7.086x2.426 - 0.5x9.8x2.426squared
0 = y0 + 17.190636 - 28.8388324
y0 = height of building = 11.648m

Answer 2

Vx==10.2*cos44 *so 17.8 = 10.2 *cos 44 *t and t =2.426 s
e=-1/2*9.8 t^2 +10.2*sin 44*t ( taking level 0 at the topo of the building
Maximum height at -9.8t +10.2 sin 44 = 0 so t =0.723 s
The rock is on the top after 2*0.723 =1.446 s
h= -1/2*9.8 (2.426^2-1.446^2)+10.2 sin 44 (2.426-1.446)=11.65m