- #1
lilstar
- 8
- 0
Hello,
I am not completely sure how to attempt this problem but I have given it a shot. If anyone knows how to do this please look it over, or post your attempt. Thank you!
An object was shot at an angle of 49 degrees and landed 60m away at the same height. The initial velocity was 23m/s.
WHat is:
The length of time that the object was in the air
The maximum height it reached
d=vt
d=(u+v/2)t
horizontal
v=?
d=60m
t=?
vertical
u=23m/s
v=-23m/s
a=-9.8m/s^2
d=0m
t=?
t=v-u/a
t=-23m/s-23m/s/-9.8m/s^2
t=4.7s
v=d/t
v=60m/4.7s
v=13m/s
v=13m/s
d=30m/s
t=?
t=d/v
t=2.3s
u=23m/s
v=0m/s
a=-9.8m/s^2
d=?
t=2.3s
d=(u+v/2)t
d=(23+0/2)(2.3)
d=26m
so the maximum height would be 26m at 2.3s?
I am not completely sure how to attempt this problem but I have given it a shot. If anyone knows how to do this please look it over, or post your attempt. Thank you!
Homework Statement
An object was shot at an angle of 49 degrees and landed 60m away at the same height. The initial velocity was 23m/s.
WHat is:
The length of time that the object was in the air
The maximum height it reached
Homework Equations
d=vt
d=(u+v/2)t
The Attempt at a Solution
horizontal
v=?
d=60m
t=?
vertical
u=23m/s
v=-23m/s
a=-9.8m/s^2
d=0m
t=?
t=v-u/a
t=-23m/s-23m/s/-9.8m/s^2
t=4.7s
v=d/t
v=60m/4.7s
v=13m/s
v=13m/s
d=30m/s
t=?
t=d/v
t=2.3s
u=23m/s
v=0m/s
a=-9.8m/s^2
d=?
t=2.3s
d=(u+v/2)t
d=(23+0/2)(2.3)
d=26m
so the maximum height would be 26m at 2.3s?