Question

A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at the same speed from the same windowsill. How long after the first ball lands does the second ball land

Answer 1

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

$t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s$

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

$t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 = \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s$

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

$t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s$

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s -  2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.