Question

A baseball is thrown straight upward with a speed of .
a. How long will it rise? [3]
b. How high will it rise? [3]
c. How long after it leaves the hand will it return to the starting point?[5]

Answer 1

Solution.

$v;$

a. $g_y=\dfrac{v_y-v_{0y}}{t}\implies t=\dfrac{vy-v_{0y}}{g_y};$

$v_y=0, t=\dfrac{-v_{0y}}{g_y};$

b. $h_y=\dfrac{vy^2-v_{0y}^2}{2g_y};$

$v_y=0, h_y=\dfrac{-v_{0y}^2}{2g_y};$

c. $t_{max}=2\sdot(-\dfrac{v_{0y}}{g_y})=-\dfrac{2v_{0y}}{g_y};$

Answer: a. $t=\dfrac{-v_{0y}}{g_y};$

b. $h_y=\dfrac{-v_{0y}^2}{2g_y};$

c. $t_{max}=-\dfrac{2v_{0y}}{g_y};$