an arrow is shot vertically upward 2 seconds later it is at a height of 42m with what velocity was the arrow shot, how long is the arrow in flight altogether
h = Vi t - 4.9 t^2
so
42 = 2 Vi -4.9*4
solve for Vi
then when is h = 0 again?
0 = Vi t - 4.9 t^2
0 = t (Vi - 4.9 t)
t = 0 at start of course
then it comes back to ground at
t = Vi/4.9
To find the initial velocity of the arrow and the total time it is in flight, we can use the equations of motion for vertically upwards projectile motion.
First, let's find the initial velocity of the arrow:
Since we know the height the arrow reaches after 2 seconds is 42m, we can use the following equation:
h = ut + (1/2)gt^2
Where:
h = height (42m)
u = initial velocity
t = time (2s)
g = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
Let's substitute the known values into the equation:
42 = u * 2 + (1/2) * (-9.8) * (2^2)
Simplifying the equation:
42 = 2u - 19.6
2u = 42 + 19.6
2u = 61.6
u = 61.6 / 2
u = 30.8 m/s
So, the initial velocity of the arrow was 30.8 m/s.
Next, let's find the total time the arrow is in flight:
To calculate the total time of flight, we can use the equation:
t = 2u / g
Substituting the values, we get:
t = 2 * 30.8 / 9.8
t = 61.6 / 9.8
t ≈ 6.29 seconds (rounded to two decimal places)
Therefore, the arrow is in flight for approximately 6.29 seconds.
In summary, the arrow was shot with an initial velocity of 30.8 m/s, and it remains in the air for approximately 6.29 seconds.