Question

an object is thrown directly up (positive direction) with a velocity (v₀) of 20.0 m/s and d₀= 0. how high does it rise (v= 0 cm/s at top of rise). remember, acceleration is -9.80 m/s².
d = ______ m
○ 40.8
○ 1.02
○ 2.04
○ 20.4

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation \( v^2 = v_0^2 + 2a(d - d_0) \), where \( v = 0 \, \text{m/s} \) (final velocity at the top), \( v_0 = 20.0 \, \text{m/s} \) (initial velocity), \( a = -9.80 \, \text{m/s}^2 \) (acceleration due to gravity), and \( d_0 = 0 \, \text{m} \) (initial position).

Step2: Rearrange the equation to solve for \( d \)

Starting with \( v^2 = v_0^2 + 2a(d - d_0) \), substitute \( v = 0 \) and \( d_0 = 0 \):
\[
0 = (20.0)^2 + 2(-9.80)(d - 0)
\]
Simplify:
\[
0 = 400 - 19.6d
\]
Then, solve for \( d \):
\[
19.6d = 400
\]
\[
d = \frac{400}{19.6} \approx 20.4 \, \text{m}
\]

Answer:

20.4 (corresponding to the option "20.4")