an arrow is shot vertically upward after 2s it is at a height of 36m although this is not necessarily the highest point with what velocity was the arrow shot what is the maximum height reached by the arrow how long is the arrow in flight altogether
the average velocity is
... 36 m / 2 s = 18 m/s
gravity has slowed decelerated) the arrow by ... 9.8 m/s² * 2 s = 19.6 m/s
[v + (v - 19.6)] / 2 = 18
... v = 27.8 m/s (launch velocity)
time to peak is
... 27.8 m/s / 9.8 m/s²
peak height is
... time to peak * 27.8 / 2
flight time is
... time to peak * 2
To find the initial velocity of the arrow, you can use the equation of motion for displacement in the vertical direction:
s = ut + (1/2)gt^2
In this case, the initial height (s) is 0m (since the arrow was shot from the ground), the time (t) is 2s, and the acceleration due to gravity (g) is approximately 9.8 m/s^2 (assuming no air resistance). Plugging in these values, we can solve for the initial velocity (u).
0 = u*2 + (1/2)(9.8)(2)^2
0 = 2u + (1/2)(9.8)(4)
0 = 2u + 19.6
Rearranging the equation to isolate u:
2u = -19.6
u = -9.8 m/s
Therefore, the arrow was shot upward with an initial velocity of 9.8 m/s.
To find the maximum height reached by the arrow, we can use the equation for maximum height in projectile motion:
h = (u^2)/(2g)
Plugging in the values, we get:
h = (9.8^2)/(2 * 9.8)
h = 9.8 m
Therefore, the maximum height reached by the arrow is 9.8 meters.
To find the total time of flight, we need to consider both the upward and downward journey of the arrow. Since the upward and downward motion of the arrow is symmetrical, the total time of flight is double the time taken to reach the maximum height.
Therefore, the total time of flight is:
2 * 2s = 4s
Thus, the arrow is in flight for a total of 4 seconds.