- #1
kathy
- 5
- 0
Problem:
Jamilla throws a stone horizontally off a pier into the ocean at a velocity of 20 m/s. If the stone is 40.0m from the edge of the pier when it hits the water, how high above the water's surface was the stone when Jamilla threw it?
Related Equations:
displacement = velocity x time
displacement = (v1) (t) + (a) (t2) /2
t= square root of (2) (d) / a
Solution:
t= square root of (2) (d) /a
t= square root of (2) (40.0m) /9.8m/s2
t= square root of 80m/ 9.8m/s2
t= square root of 8.16
t= 2.9s
d= (v)(t)
d= (20m/s) (2.9s)
d= 58.0m
t= time
d= displacement
a= acceleration
Having trouble with the equations again. I want to know whether the equation I'm using is right or do I have to use a different equation?
Therefore the stone was 58.0m high above the ground when Jamilla threw it.
Jamilla throws a stone horizontally off a pier into the ocean at a velocity of 20 m/s. If the stone is 40.0m from the edge of the pier when it hits the water, how high above the water's surface was the stone when Jamilla threw it?
Related Equations:
displacement = velocity x time
displacement = (v1) (t) + (a) (t2) /2
t= square root of (2) (d) / a
Solution:
t= square root of (2) (d) /a
t= square root of (2) (40.0m) /9.8m/s2
t= square root of 80m/ 9.8m/s2
t= square root of 8.16
t= 2.9s
d= (v)(t)
d= (20m/s) (2.9s)
d= 58.0m
t= time
d= displacement
a= acceleration
Having trouble with the equations again. I want to know whether the equation I'm using is right or do I have to use a different equation?
Therefore the stone was 58.0m high above the ground when Jamilla threw it.