An arrow is shot straight upward into the air and it returns to the ground with a speed of 28m/s. What was the maximum height reached by the arrow?
v = Vi - 9.81 t
if it came down at 28 m/s, it started up at 28 m/s (symmetry)
v = 28 - 9.81 t
at the top, v = 0
0 = 28-9.81 t
so
t = 28/9.81 at the top, same time falling
h = (1/2)(9.81) t^2
h = (4.9)(28/9.81)^2
h = 4.9
To find the maximum height reached by the arrow, we need to use the principles of projectile motion. When an object is shot straight upward and then returns to the ground, the maximum height reached will be at the midpoint of the upward and downward journey.
Let's break down the problem into steps:
Step 1: Find the initial velocity of the arrow when it was shot upward.
Since there is no information given about the initial velocity, we'll assume that it was the same as the final velocity when it returns to the ground. Therefore, the initial velocity of the arrow is also 28 m/s.
Step 2: Calculate the time it takes for the arrow to reach the highest point.
To find this, we can use the fact that the velocity at the highest point is zero. The vertical component of motion can be described by the equation:
v = u + gt
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (28 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken
Substituting the values into the equation, we get:
0 = 28 - 9.8t
9.8t = 28
t = 28 / 9.8
t ā 2.86 seconds
Step 3: Find the maximum height reached.
The height can be determined using the formula for vertical displacement:
s = ut + (1/2)gt^2
where:
s = displacement or height
u = initial velocity
t = time taken
g = acceleration due to gravity
Substituting the values, we have:
s = (28)(2.86) + (1/2)(-9.8)(2.86^2)
s = 80.08 - 41.3328
s ā 38.75 meters
Therefore, the maximum height reached by the arrow is approximately 38.75 meters.
To determine the maximum height reached by the arrow, we can use the kinematic equation for vertical motion.
The equation is:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (0 m/s when the arrow reaches its maximum height)
- vi is the initial velocity (the velocity at the highest point is also 0 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2, assuming downward direction as negative)
- d is the vertical displacement (maximum height)
Rearranging the equation:
0^2 = 28^2 + 2(-9.8)d
0 = 28^2 - 19.6d
-784 = -19.6d
Solving for d:
d = -784 / -19.6
d ā 40 meters
Therefore, the maximum height reached by the arrow is approximately 40 meters.