A 3.60 kg ball is dropped from the roof of a building 184.8 m high. While the ball is falling to earth, a horizontal wind exerts a constant force of 12.1 N on the ball.
(acceleration of gravity = 9.81 m/s^2)
How far from the building does the ball hit the ground?
What is the speed when it hits the ground?
http://www.jiskha.com/display.cgi?id=1257565507
the acceleration is 6.1380649213407
dont know the rest its a 3 part problem
Read my first answer. Bob Pursley gave you the link to it.
You were not asked to compute an acceleration.
You only asked a one part question
yes
To find the distance from the building where the ball hits the ground, we can use the equation of motion under constant acceleration:
d = v₀t + 0.5at^2
Where:
- d is the distance from the building where the ball hits the ground
- v₀ is the initial velocity of the ball (0 m/s since it is dropped)
- t is the time it takes for the ball to hit the ground
- a is the acceleration due to gravity (-9.81 m/s^2)
First, let's find the time it takes for the ball to hit the ground. We can use the equation of motion:
v = v₀ + at
Since the ball is dropped, the initial velocity v₀ is 0 m/s. We can rearrange the equation to find the time:
t = -v₀ / a
Plugging in the values, we have:
t = 0 m/s / (-9.81 m/s^2)
t = 0 s
Since the time is 0 seconds, the ball hits the ground instantly after dropping. Therefore, the distance from the building where the ball hits the ground is 0 meters.
Now let's find the speed at which the ball hits the ground. We can use the equation of motion:
v = v₀ + at
Again, since the ball is dropped, the initial velocity v₀ is 0 m/s. Therefore, the equation simplifies to:
v = at
Plugging in the values, we have:
v = 9.81 m/s^2 * 0 s
v = 0 m/s
Therefore, the speed at which the ball hits the ground is 0 m/s.