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.
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
5 answers
Upwards
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
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
How did you get three? .-.