Question

an object is thrown directly up (positive direction) with a velocity (v₀) of 20.0 m/s and d₀= 0. determine how long it takes to get to the maximum height of 20.4 m. 4.16 2.04 5.00

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation for vertical motion: \( d = d_0 + v_0 t - \frac{1}{2} g t^2 \), where \( d \) is the final displacement, \( d_0 \) is the initial displacement, \( v_0 \) is the initial velocity, \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)), and \( t \) is time. Given \( d_0 = 0 \), \( v_0 = 20.0 \, \text{m/s} \), \( d = 20.4 \, \text{m} \), the equation becomes \( 20.4 = 0 + 20.0t - \frac{1}{2}(9.8)t^2 \).

Step2: Simplify the equation

Simplify the equation: \( 20.4 = 20t - 4.9t^2 \). Rearrange to standard quadratic form: \( 4.9t^2 - 20t + 20.4 = 0 \).

Step3: Solve the quadratic equation

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b = -20 \), \( c = 20.4 \).

First, calculate the discriminant: \( \Delta = (-20)^2 - 4(4.9)(20.4) = 400 - 4(4.9)(20.4) \)

\( 4(4.9)(20.4) = 4 \times 4.9 \times 20.4 = 19.6 \times 20.4 = 400 - 399.84 = 0.16 \)

Then, \( t = \frac{20 \pm \sqrt{0.16}}{2(4.9)} = \frac{20 \pm 0.4}{9.8} \)

We have two solutions:

• \( t_1 = \frac{20 + 0.4}{9.8} = \frac{20.4}{9.8} \approx 2.08 \) (close to 2.04, considering rounding differences)
• \( t_2 = \frac{20 - 0.4}{9.8} = \frac{19.6}{9.8} = 2.0 \) (not in options, but the closest option is 2.04, likely due to rounding in \( g \) or calculation)

Alternatively, at maximum height, the final velocity \( v = 0 \). Use \( v = v_0 - gt \), so \( 0 = 20 - 9.8t \), solving for \( t \): \( t = \frac{20}{9.8} \approx 2.04 \, \text{s} \).

Answer:

2.04