Can someone help me with a physics problem?a) A 3.90 kg ball is dropped from the roof of a building 172.6 m high. While the ball is falling to...

a)

Since the wind force is
horizontal, it has no vertical component.

Hence the verical
acceleration is entirely due to the gravity force. Also the vertical component of velocity is
also entirely due to gravitation.

Using the equation of motion for
the falling body, the vertical distance s covered  is given by:

s =
ut +(1/2)gt^2, where u = initial speed of the ball. u= is zero as the ball is dropped from a
height. s = 171.6 m ,  g = 9.81 m/s^2. We determine the time t with the given
details.

171.6 =
0t+(1/2)*9.81*t^2.

171.6*
(1/2)*9.81*t^2.

t^2 = 2*171.6/9.81.

t =
sqrt(2*171.6/9.81) = 5.91 sec.

b)

There
is no horizontal component of the graviational force which is entirely vertical. So only the
constan horizontal force of 13.1N of force by wind  generate an acceleration a = Forc/mass of the
ball = 13.1N/3.91 Kg = 3.5666.. m/s^2 in the ball.

So the equation
of motion for the horizontal direction is also:

s = ut+(1/2)at^2,
where s is the distance covered and  t is the time the ball is in air till it hits the ground and
a = 13.1/3.9 = 3.3589....m/s^2. t = 5.9148 second as we got in a.

s
= 0* 5.91+(1/2) (13.1/3.9)*(5.91..)^2

s =  58.76 meter away from the
building.

c)

The speed (or rather
velocity) of the ball while it hits the ground is due to both vertical as well as horizontal
velocities.

Vertical vecity v = u+gt = 0+9.81 *5.9148 = 58.0242 m/s
vertically down.

Horizontal velocity = u+at = 0+(13/3.9)(5.9148) =
19.8677

Therefore the combined velocity of the ball =
sqrt(horizontal vel^2+vertical vel ^2) = sqrt{58.0242^2+19.8677^2) =  61.3313m/s^2. The direction
of the velocity is arc tan (58.0242/19.8677) = 71 degree below
horizontal.