An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm/s at top of rise). Remember, acceleration is -9.80 m/s2.
21m
To find out how high the object rises, we can use the kinematic equation:
v^2 = vo^2 + 2a(d - do)
Where:
v = final velocity (0 m/s at the top)
vo = initial velocity (20.0 m/s)
a = acceleration (-9.80 m/s^2)
d = final position (height it rises to)
do = initial position (0 m)
First, let's solve for d:
0^2 = (20.0)^2 + 2(-9.80)(d - 0)
Simplifying the equation, we get:
0 = 400 - 19.6d
Rearranging the equation to solve for d:
19.6d = 400
d = 400 / 19.6
d ≈ 20.41 meters
Therefore, the object rises to approximately 20.41 meters.
To determine the height the object rises, we can use the kinematic equation for vertical motion:
vf^2 = vo^2 + 2aΔd
where:
- vf is the final velocity (v = 0 m/s at the top of the rise),
- vo is the initial velocity (20.0 m/s),
- a is the acceleration (-9.80 m/s^2),
- Δd is the displacement (height we want to find).
Rearranging the equation, we have:
Δd = (vf^2 - vo^2) / (2a)
Substituting the given values into the equation:
Δd = (0^2 - 20.0^2) / (2*-9.80)
Before we calculate this, let's simplify further:
Δd = (-400) / (-19.60)
Δd = 20.41…
The object rises to a height of approximately 20.41 meters.