Need Help with physics and a ball that falls from a building

Physics
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Dec 1st, 2015

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It is given that the ball is thrown form the Empire state building so the time taken to reach the ground is 4.27 seconds.

so,

Time taken (t) = 4.27 s

The Initial Velocity of the ball (u) = 0 m/s

a) There will not be any velocity and acceleration of the ball in horizontal direction. As the ball is thrown from a height, it falls freely from the empire state building to the ground.

b) The ball experiences a velocity and acceleration in the vertical direction.

VELOCITY (VERTICALLY DOWNWARD)

The ball experiences a velocity in vertically downward direction and this velocity is dependent on time. From the basic laws of motion.

we know that,

final velocity(v) = initial velocity(u) + acceleration(a)*time(t)

v=u+at

here, u = 0

thus,

v = at

here,

a = -g = -9.81 m/s^2

the negative sign indicates the direction of the velocity and in this case it is vertically downward.

ACCELERATION (VERTICALLY DOWNWARD)

As the ball is falling down it is acted  by only gravitational force of the earth. So, the ball experiences an acceleration due to gravity which is 9.81 m/s^2. and this acts in vertically downward direction.

c) Let the distance traveled by the ball be "s" (i.e. from the top of building to the ground)

From basic laws of motion,

we know that

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

here, u = 0 , a = 9.81 m/s^2 , t = 4.27 s

s = (1/2)*a*(t)^2 = (1/2)*9.81*(4.27)^2 = 89.432 m.

d) The velocity of the ball can be found by the following

we know that,

final velocity(v) = initial velocity(u) + acceleration(a)*time(t)

v=u+at

here, u = 0

thus,

v = at

here, a = 9.81m/ s^2 and t = 4.27s

so the velocity of the ball before hitting the ground is

v = 9.81*4.27 = 41.88 s


Please let me know if you need any clarification. I'm always happy to answer your questions.
Dec 1st, 2015

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