Asked by Jack
A stone is thrown from a height of 1.8 with a velocity of 29.4. How long does it take to reach its greatest height?
I'm sure it has something to do with the s=ut+1/2 at squared not sure though
I'm sure it has something to do with the s=ut+1/2 at squared not sure though
Answers
Answered by
Damon
Thrown at what angle to horizontal?
In furlongs / month?
It has to do with
distance = initial position + Vi t + (1/2) a t^2
but that may not be the easy way to do it.
conservation of energy is faster if the max height is all you need.
In furlongs / month?
It has to do with
distance = initial position + Vi t + (1/2) a t^2
but that may not be the easy way to do it.
conservation of energy is faster if the max height is all you need.
Answered by
Jack
Upwards
Answered by
Damon
and what units ? Yikes !
what do you use for g?
9.81 m/s^2 or 32 ft/s^2
what do you use for g?
9.81 m/s^2 or 32 ft/s^2
Answered by
Damon
if in meters/ second^2
v = Vi - 9.81 t
at the top v = 0
so
0 = 29.4 - 9.81 t
t = 3 seconds drifting upawrds
Now you could do the 4.9 t^2 thing but it
is easier just to use the average speed up
which is (1/2)29.4 m/s = 14.7 m/s
height = 1.8 m + 14.7 m/s * 3 seconds
= 1.8 + 44.1
= 45.9 meters total height
v = Vi - 9.81 t
at the top v = 0
so
0 = 29.4 - 9.81 t
t = 3 seconds drifting upawrds
Now you could do the 4.9 t^2 thing but it
is easier just to use the average speed up
which is (1/2)29.4 m/s = 14.7 m/s
height = 1.8 m + 14.7 m/s * 3 seconds
= 1.8 + 44.1
= 45.9 meters total height
Answered by
Jack
How did you get three? .-.