You throw an object up with a speed of v0y = 9 m/s from a height of y = 25 m. a) How long, in seconds, does it take for the object to reach the ground?

b) What is the object's final velocity, in meters per second, as it impacts the ground?

c) Find in the time, in seconds, if you instead threw the object down the same velocity, Voy

a) 3.36 sec

b) 23.9 m/s

c) 1.52 sec

Explanation:

maximum height

H = 25 +v_y^2/2g

H=25+9^2/2*9.81 =29.13 m

time taken to reach maximum height = t_1 =v_y/g = 9/9.81 =0.92 seconds

time taken to fall back to ground from maximum height = √(2gH)/g

= √(2*9.81*29.13)/9.8 = 2.439 seconds

a) Total time taken to reach ground T = t_1+t_2 =0.92+2.439=3.36 sec

b) objects final speed = √2gH = 23.9 m/s

c) let total time be t then

25 = 9t +0.5gt^2

Solving we get t= 1.52 seconds

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000

there are no atoms