Marie shoots an arrow vertically upward at a sandhill crane (she misses the crane, so you PETA members can relax). If the arrow travels to a maximum height of 99.5 meters, calculate how fast it was shot.
And, for how much time was Marie's arrow in the air?
a. Vf^2 = Vo^2 + 2g*d,
Vo^2 = Vf^2 - 2g*d,
Vo^2 = 0 - (-19.6)99.5 = 1950,
Vo = 44.2 m/s.
b. Tr=Tf= (Vf-Vo)/g=(0-44.2)/-9.8=2.26s
= Rise time = Fall time.
Tr + Tf = 2.26 + 2.26 = 4.52 s. = Time in air.
To calculate the speed at which Marie's arrow was shot, we can use the equation for projectile motion:
v = √(2gh)
Where:
v is the initial velocity of the arrow,
g is the acceleration due to gravity (approximately 9.8 m/s²), and
h is the maximum height reached by the arrow (99.5 meters).
Plugging in the values:
v = √(2 × 9.8 × 99.5)
v ≈ √(1963)
v ≈ 44.27 m/s
So, the arrow was shot at approximately 44.27 m/s.
To find out how much time the arrow was in the air, we can use the equation for time of flight:
t = √(2h / g)
Plugging in the values:
t = √(2 × 99.5 / 9.8)
t ≈ √(202 / 9.8)
t ≈ √(20.6122)
t ≈ 4.54 seconds
Therefore, Marie's arrow was in the air for approximately 4.54 seconds.
To calculate the initial speed at which Marie’s arrow was shot, we can use the kinematic equation for vertical motion:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s at the maximum height)
u = initial velocity (what we are trying to find)
a = acceleration due to gravity (approximately 9.8 m/s^2)
s = displacement (in this case, the maximum height, which is 99.5 meters)
Since the final velocity is zero at the maximum height, the equation becomes:
0 = u^2 + 2(9.8)(99.5)
Rearranging the equation, we can solve for u:
u^2 = -2(9.8)(99.5)
u^2 = -1942.1
u = √(-1942.1) (taking the square root of both sides)
Since the square root of a negative number is not a real number, we conclude that there must be an error in the given values or the question itself. The initial speed cannot be determined based on the given information.
As for the time Marie's arrow was in the air, we can use the kinematic equation:
v = u + at
At the maximum height, the final velocity is 0 m/s. Given that the acceleration due to gravity is -9.8 m/s^2 (as the arrow is moving against gravity), we can rewrite the equation as:
0 = u - (9.8)t
Rearranging the equation, we can solve for t:
t = u / 9.8
However, since we don't have the initial velocity (u), we cannot determine the time the arrow was in the air without additional information.