How high a building could Superman jump over if he were to leave the ground with a speed of 60.0 m/s at an angle of 75°?
Answers
Answered by
0
171.04
m
Explanation:
The vertical component of his velocity is given by:
V
y
=
60
cos
(
90
−
75
)
=
60
sin
(
75
)
=
57.95
m/s
We can use:
v
2
=
u
2
+
2
a
s
∴
0
=
57.9
2
+
2
(
−
9.8
)
s
∴
s
=
171.04
m
Similar questions