A stone is thrown horizontally at a speed of 8.0m/s from the edge of a cliff 80m in height. How far from the base of the cliff will the stone strike the ground?
Answers
GOOD MORNING HRE IS YOUR ANSWER
Answer and Explanation:
The given value in the problem is the initial speed of the stone
v
0
=
8.0
m/s
, the initial height of the stone
y
0
=
80
m
and the launching angle
θ
0
=
0
∘
, this is because the stone is thrown horizontally. Let's start by getting the time when the stone hits the ground
y
t
=
0
m
. This is computed through the position equation along the vertical.
y
t
=
y
0
+
v
0
sin
(
θ
0
)
t
−
1
2
g
t
2
0
m
=
80
m
+
(
8.0
m/s
)
sin
(
0
∘
)
t
−
1
2
(
9.8
m/s
2
)
t
2
0
=
80
m
+
(
8.0
m/s
)
(
0
)
t
−
(
4.9
m/s
2
)
t
2
(
4.9
m/s
2
)
t
2
=
80
m
+
(
0
)
t
(
4.9
m/s
2
)
t
2
=
80
m
+
0
t
2
=
80
m
(
4.9
m/s
2
)
t
2
=
16.3265
s
2
√
t
2
=
√
16.3265
s
2
t
=
4.04
s
Substitute this to the position equation along the horizontal to get the range of the stone or the distance from the base of the cliff to the point where the stone struck the ground.
x
t
=
x
0
+
v
0
cos
(
θ
0
)
t
x
t
=
0
m
+
(
8.0
m/s
)
cos
(
0
∘
)
(
4.04
s
)
x
t
=
0
+
(
8.0
m/s
)
(
1
)
(
4.04
s
)
x
t
≈
32
m