Basically, I don't have much idea how to solve this free fall problems. Please help me.?

Question

I need the complete solution and the formulas used which I can study. #1.A piece of stone is dropped from the rooftop of the buliding that is 12m high. a.) How long does the stone take to reach the ground. b.) With what speed does the stone hit the ground? #2 A stone is catapulted vertically upwards from the ground with a speed of 25m/s. a.) How high does the stone rise? b.) How long does the stone take up to rise up and to return to the ground? c.) With what speed does the stone hit the ground? #3. A ball is dropped from a window 64ft above the ground. a.) How long does it take the ball to reach the ground? b.) What is its final velocity? #4. A ball is thrown downward with an initial velocity of 20ft/s. a.)How fast is it moving 2 sec later? b.) How far does it fall in these two seconds? #5. A ball is thrown upward with an initial velocity of 20m/s. a.) How fast is it moving 1/2 seconds later and in what direction?. b.) How fast is it moving 2 seconds later and in what direction?

Answer 1

#1  

A) d = .5at^2 + vi(t)  (vi is the initial velocity, which is zero in this case)

-12 = .5(-9.8)(t^2)

t = 1.6 s

B) vf = at + vi (vi is the initial velocity, which is zero in this case)

vf = -9.8 * 1.6 = 15 m/s

#2

A)  vf^2 = vi^2 + 2ad  

At the highest point, the final velocity will be zero.

0 = 25^2 + 2(-9.8)(d)

-625 = -19.6d

625/19.6 = d = 32 m

B)  Since it will go the same distance up as it goes down, the change in distance, d, will be zero.

d= 0 = .5at^2 +25t

t(.5at + 25) = 0

.5(-9.8)t = -25

t = 5.1 s

C) vf^2 = vi^2 + 2ad

If we start at the highest point, vi = 0.

D at the highest point was 32 m

vf^2 = 2(9.8)(32)

Vf = -25 m/s

It is negative because it is going in the downward direction, which is conventionally notated as negative.

#3

A) d = .5at^2 +vi(t)

a = 9.8 m/sec^2 * 39.37/12 ft per meter = 32.152 ft per sec^2

-64 = .5(-32.152)t^2

t = 4.0 s

B) vf^2 = vi^2 + 2ad

vf^2 = 2(-32.152)(-64)

vf = -64 ft/s

#4

A)  vf = at + vi

vf = -32.152(2) + (-20) = -84 ft/s

B)  d = .5at^2 + vi(t)

d = .5(-32.152)(2^2) + (-20)(2) = -104 m

#5

A)  In the direction it was thrown, it is still moving 20 m/s.  In addition it will be moving 9.8 * .5 (a*t) = 4.9 m/s downward, or -4.9 m/s.  

B)  It will still be moving 20 m/s in the direction it was thrown. In addition it will be moving 9.8 * 2 (a*t) = 19.6 m/s downward, or -19.6 m/s.