an arrow is shot vertically upward 2 seconds later it is at an height of 42m (not necessarily the highest point) with what velocity was the arrow shot how long was the arrow in flight altogether
average velocity = distance / time =
42 m / 2 s = 21 m/s
average velocity =
(initial + current) / 2
gravity slowed the arrow for 2 sec
... current = initial - (2 * 9.8)
(Vi + Vi - 19.6) / 2 = 21 m/s
solve for Vi
time up = time down
flight time = 2 (Vi / g)
To determine the initial velocity and total time in flight of the arrow, we can use the equations of motion for vertical motion.
Let's consider the following variables:
- Initial velocity (u)
- Final velocity (v)
- Acceleration due to gravity (g) = -9.8 m/s² (since the arrow is moving upward)
- Time taken to reach a height of 42m (t)
We can use the kinematic equation for vertical motion:
v = u + gt
Given that the final velocity at a height of 42m is 0 m/s (when the arrow reaches the highest point of its trajectory before falling back down), we have:
0 = u - 9.8 * 2
Solving this equation, we can calculate the initial velocity (u):
u = 9.8 * 2
u = 19.6 m/s
Therefore, the arrow was shot vertically upward with an initial velocity of 19.6 m/s.
To find the total time in flight, we need to account for both the upward and downward journeys of the arrow. The time taken to reach the highest point is referred to as the "time of ascent," and the time taken to fall back to the starting height is referred to as the "time of descent."
The time of ascent can be calculated using the equation:
t = (v - u) / g
Given that the final velocity at the highest point is 0 m/s and the initial velocity is 19.6 m/s, we have:
t = (0 - 19.6) / -9.8
t = 2 seconds
Since the arrow took 2 seconds to reach the highest point, the time of descent will also be 2 seconds.
The total time in flight of the arrow is the sum of the time of ascent and the time of descent:
Total time = Time of ascent + Time of descent
Total time = 2 + 2
Total time = 4 seconds
Therefore, the arrow was in flight for a total of 4 seconds.