an archer shoots an arrow straight up into the air with an initial velocity of 45 m/s (U). calculate the maximum height of the arrow
To calculate the maximum height of the arrow, we can use the equations of motion.
The key information we have is the initial velocity of the arrow, which is 45 m/s. Since the arrow is shot straight up, we can assume the acceleration due to gravity acts in the opposite direction, which is -9.8 m/s^2 (taking downward as the positive direction).
We can use the following equation to find the maximum height (H) reached by the arrow:
v^2 = u^2 + 2as
where:
v = final velocity (which is 0 m/s at the maximum height)
u = initial velocity (45 m/s)
a = acceleration (-9.8 m/s^2)
s = displacement
Rearranging the equation, we have:
0^2 = (45 m/s)^2 + 2(-9.8 m/s^2)s
Simplifying the equation further, we get:
0 = 2025 m^2/s^2 - 19.6 m/s^2 * s
Solving for 's', we have:
19.6 m/s^2 * s = 2025 m^2/s^2
s = 2025 m^2/s^2 ÷ 19.6 m/s^2
s ≈ 103.57 m
Therefore, the maximum height reached by the arrow is approximately 103.57 meters.
To calculate the maximum height of the arrow, we need to consider the motion of the arrow under the force of gravity.
The vertical motion of the arrow is described by the equation:
h = (U^2) / (2g)
where h is the maximum height, U is the initial vertical velocity, and g is the acceleration due to gravity.
In this case, the initial vertical velocity U is given as 45 m/s.
The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Now, we can substitute the given values into the equation to calculate the maximum height:
h = (45^2) / (2 * 9.8)
= 2025 / 19.6
≈ 103.57 meters
Therefore, the maximum height of the arrow is approximately 103.57 meters.
Conservation of energy applies.
Initial KE= final PE
1/2 m v^2=mgh (at the max height, velocity is zero)
h= v^2/g or 45^2/((2)(9.8)) that is pretty high.