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
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.
Upwards
and what units ? Yikes !
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
How did you get three? .-.
You're on the right track! To find the time it takes for the stone to reach its greatest height, you can use the equation for displacement:
s = ut + (1/2)at²
In this case, the initial velocity (u) is 29.4 m/s, the acceleration (a) is -9.8 m/s² (assuming upwards as positive), and the initial displacement (s) is 1.8 m. We'll be solving for time (t).
However, we'll need to consider that at the highest point of the stone's trajectory, the velocity becomes 0. This means that the final velocity (v) is 0 m/s. With this information, we can rearrange the equation as:
v² = u² + 2as
Since v is 0 at the highest point, we can rewrite this equation as:
0 = u² + 2as
Rearranging again:
u² = -2as
Now, we substitute the values we have:
(29.4)² = -2 × (-9.8) × 1.8
Simplifying:
861.36 = 35.28
This equation is not possible, and it's because we've assumed a negative acceleration (gravity) but used a positive displacement. In reality, the displacement at the highest point of the stone's trajectory should be negative.
To find the time it takes for the stone to reach its greatest height, it's important to know the proper initial displacement value. Please double-check the information and provide the correct initial displacement (s).