Ball Being Dropped

Question

At time t=0 a ball is dropped from a height of 100 feet. One second later, another ball is dropped from the height of 75 feet. The position of the ball is given by the formula:

s(t) = -16t^2 + Vot + So

, where s is measured in feet and t is measured in seconds.

a) Which ball hits the ground first?

b) Calculate speeds of the impacts.

c) Calculate acceleration of the first ball at t=1.

Answer 1

The formula for gravity/distance is (t^2+t)/2*32.2.

To fall 100' it'll take 2.042 sec.

The first dropped ball will hit the ground first.

The speed of anything dropped is 32.2'/sec^2; meaning:

Dropped; 1 sec, 32.2'.

Sec #2; ball is now falling 64.4'/sec.

#3 96.6 "

#4 128.8 ".......................

So use the formula above.

Answer 2

Step 1: draw a picture. Since I am assuming that the balls are dropped from rest (not stated), v0 = 0. Both balls fall to the ground, so s(t final) = 0 for both.

a) solve your position equation for time: this gives the time of fall for each ball

0 = -16t^2 + 0t + 100 for the first ball ==> t = √(100/16) = 2.5 s

0 = -16t^2 + 0t + 75 for the second ball ==> t = √(75/16) = 2.16 s

the times of fall aren't the complete answer to which hits first. To compare the times, have to add 1 second to the time of the second ball, because it was dropped one second after the first was dropped.

So the ball dropped from 100 ft lands first.

b) the velocity is ds/dt, so differentiate the position equation: v(t) = -32 t + v0 = -32 t

To determine the velocities when the balls hit, use the actual times of fall, calculated in a).

v1 = -32 (2.5) = -80 ft/s

v2 = -32 (2.16) = - 69 ft/s

c) the acceleration is the second derivative of position, d²s/dt² , so differentiate the position equation twice. a = -32 ft/s², a constant value.

Answer 3

initial velocity for both balls is zero

so the equation for height becomes

h(t) = -16t^2 +100=0

for the first ball

and h(t)= -16t^2 + 75 =0 for the 2nd ball

solve for t

and the first ball hits ground in 5/4 = 1.25 seconds

with the 2nd ball landing in (5/4)sqr3 seconds

(5/4)sq3 = about (5/4)(1.732) = 2.165 seconds

Also add another second to the 2nd ball's time to hit ground as it was dropped 1 second later

From t=0 to t= 3.165 for the 2nd ball shows it landing nearly 2 seconds after the 1st ball.

If you're tempted to subtract 1 second, 2.165-1 = 1.165, that is less than 1.25 seconds, but

to calculate which landed first, you need to add 1 second, and 3.165>1.25 seconds

IF both balls had been dropped at the same time, the 2nd ball would have hit ground first.