Answer to Question #172408 in Classical Mechanics for Jethro Torio

Question

A ball is thrown vertically upward with a velocity of 30m/s^-1

Give its velocity after 1s.

Determine the time required for the ball to reach max height.

Find the time required to reach a height of 30m.

Solve for the maximum height the ball achieves.

Answer 1

(a)

$$v=v_0-gt,$$ $$v(t=1\ s)=30\ \dfrac{m}{s}-9.8\ \dfrac{m}{s^2}\cdot1\ s=20.2\ \dfrac{m}{s}.$$

(b)

$$v=v_0-gt,$$ $$0=v_0-gt,$$ $$t=\dfrac{v_0}{g}=\dfrac{30\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=3.06\ s.$$

(c)

$$y=v_0t-\dfrac{1}{2}gt^2,$$ $$30=30t-4.9t^2,$$ $$4.9t^2-30t+30=0.$$

This quadratic equation has two roots: $t_1=4.86\ s$ and $t_2=1.26\ s.$ The correct answer is $t=1.26\ s$ (because $t_2$ corresponds to the case when the ball returns from its maximum height to the 30 meters height).

(d)

$$y_{max}=30\ \dfrac{m}{s}\cdot3.06\ s-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(3.06\ s)^2=45.9\ m.$$