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 Earth, a horizontal wind exerts a constant force of 13.1 N on the ball. How long does it take to hit the ground?The acceleration of gravity is 9.81 m/s^2. Answer in units of s.

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The distance an object travels is related to the initial velocity of the object, how long it is falling, and the acceleration of gravity.

The general formula is: distance = v(initial)t (seconds) + 1/2 gt^2.

In your problem the initial velocity is assumed to be zero since the ball was dropped.

The distance is 172.6 m, and the value of g is given as 9.81 m/s/s.

Then:

172.6 = 0 = 1/2 * 9.81*t^2

Solving for t you get a time of 5.93 seconds.

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.

To solve this question we need to take into consideration only the gravitational force. The horizontal force exerted by the wind has no impact on the time taken by the ball to hit the ground. As no resistance to vertical movement of the ball is given it is assumed to be absent.

Also the weight of the ball has no impact on the time taken to hit the ground.

**Solution:**

We know that for a body moving under the force of gravitation:

s = ut + (1/2)g*t^2

Where:

s = distance moved = 172.6 m (Given)

u = initial velocity = 0

g = acceleration due to gravity = 9.81 m/s^2 (Given0

t = Time of travel = ?

Substituting given values in above equation:

172.6 = (1/2)*9.81*t^2

==> t^2 = 2*172.6/9.81 = 35.18858

Therefore:

t = 5.932 s

Answer:

Ball will take 5.932 s to hit the ground.

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