Question

A ball is thrown vertically upward with a speed of +18 m/s. (assume g=10 m/s2
)

a. How high does it rise?


b. How far is the horizontal distance that can be traveled?


c. How long does it take to reach its highest point?


d. How long does the ball take to hit the ground after it reaches its highest point?


e. What is its velocity when it returns to the level from which it started?

Answer 1

(a)

$$v^2=v_0^2-2gh.$$ $$0=v_0^2-2gh,$$ $$h=\dfrac{v_0^2}{2g}=\dfrac{(18\ \dfrac{m}{s})^2}{2\cdot10\ \dfrac{m}{s^2}}=16.2\ m.$$

(b) Since the horizontal component of ball's velocity equals zero (ball is thrown vertically upward), the horizontal distance traveled by the ball equals zero.

(c)

$$v=v_0-gt,$$ $$0=v_0-gt,$$ $$t=\dfrac{v_0}{g}=\dfrac{18\ \dfrac{m}{s}}{10\ \dfrac{m}{s^2}}=1.8\ s.$$

(d) The ball takes 1.8 s to hit the ground after it reaches its highest point.

(e) (e)

$$KE=PE,$$ $$\dfrac{1}{2}mv^2=mgh,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot10\ \dfrac{m}{s^2}\cdot16.2\ m}=18\ \dfrac{m}{s}.$$