A student drops a 0.4kg ball from a height a of 49m above the ground. Neglect drag. Answer each of the following questions about the ball during the flight. What is the magnitude of the net force on the ball?
What is the magnitude of the ball's acceleration?
How long does the ball take to hit the ground?
What is the ball's speed just before striking the ground?

• The only force acting on the ball, and thus the net force, is due to its own weight, with magnitude

w = m g = (0.4 kg) (9.8 m/s²) ≈ 3.9 N

pointing downward.

• The ball is in free fall, so its acceleration is g = 9.8 m/s² (also pointing downward).

• The ball is dropped from a height of 49 m, so its height y at time t is

y = 49 m - 1/2 g t ²

Set y = 0 and solve for t :

0 = 49 m - 1/2 g t ²

t ² = (98 m) / g

t = √((98 m) / g) = √(10) s ≈ 3.2 s

• The ball's velocity v at time t is

v = - g t

so that at the time found previously, the ball will have attained a velocity of

v = - g (3.2 s) ≈ -31 m/s

and thus a speed of about 31 m/s.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form