Question

3. (High Elevation Upward)

A baseball player throws a baseball straight upward with a velocity of 12.0 m/s from an initial
height of 2.2 m.
a.) How long does it take for the baseball to hit to the ground?
b.) What is the velocity of the baseball just before it hits the ground?
c.) What is the maximum height attained by the baseball?

Answer 1

Below

Step-by-step explanation:

a) d= 2.2 + 12 t + 1/2 at^2 d= 0 when it hits the ground

a = - 9.81 m/s^2

solve for t = 2.62 s ( Use Quadratic Formula or graphing method)

b) vf = vo + at

= 12 - 9.81 (2.62) = -13.7 m/s ( or 13 m/s DOWNWARD)

c) vf = 0 at max height

0 = vo + at

= 12 - 9.81 t finds t = 1.22 s

the d = 2.2 + 12 (1.22) - 1/2 * 9.81 (1.22^2) = 9.53 m ( if you graphed the equation for part a you can see this value at the peak)