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You throw a ball downward from a window at a speed of 2.0 m/s. How fast will it be moving when it hits the sidewalk 2.5 m below?

Respuesta :

Hagrid
this will require the use of kinematic equations:we have:Vstarting = 2.0 m/sHeightStarting = 2.5 mand we need to find Vfinal
the kinematic equation of V^2 = V^2 + 2ah can be used velocity final ^2 = velocity starting ^2 + 2 * acceleration of gravity * change of height
there are two terms which we are given, velocity starting and change in height, we are missing two terms: final velocity (which we are looking for) and acceleration of gravity.
acceleration of gravity is a number which is always the same, and can be found usually in the front or back cover of a physics book. we are using meters, so we will use 9.81 m/sec^2
so now we have 3 knowns, and one unknown. 
all that is left is to plug the three known values into the kinematic equation I listed above, and use algebra to solve for the one unknown, which is final velocity.

i guess it would be better to uses=ut+1/2at2

and find t

then use v=u+at

where s is the distance=2.5

u=initial speed=0

v=final speed which we have to find

a=acceleration which is due to gravity (take as positive)

t=time

v = square root(v^2+2ah)

v = square root(2^2+2∗9.8∗2.5)
v = square root(53)
v = 7.3

The speed of the ball when it is about to hit the sidewalk will be  [tex]\boxed{7.28{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}[/tex].

Further Explanation:

Given:

The initial speed of the ball is  [tex]2\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex].

The height through which the ball falls is  [tex]2.5\,{\text{m}}[/tex].

Concept:

As the ball is dropped from the window by giving it some initial velocity and it falls freely without any resistance. The motion of the ball will be freely under the acceleration due to gravity.

The final speed obtained by the ball as it reaches the sidewalk can be determined by using the third equation of motion. The equation of motion is written as:

[tex]\boxed{v_f^2 = v_i^2 + 2aS}[/tex]

 

Here, [tex]{v_i}[/tex] is the initial velocity of the ball, [tex]{v_f}[/tex] is the final velocity attained by the ball, [tex]a[/tex] is the acceleration due to gravity and [tex]S[/tex] is the distance covered by the ball.

Substitute the values in the above equation:

[tex]\begin{aligned}v_f^2&={\left( 2 \right)^2} + \left( {2 \times 9.8 \times 2.5}\right)\\{v_f}&= \sqrt {4 + 49}\\&= 7.28\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}[/tex]

Thus, the speed of the ball when it is about to hit the sidewalk will be  [tex]\boxed{7.28{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}[/tex].

Learn More:

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Motion under gravity

Keywords:  Ball downwards, from a window, speed of 2m/s, initial velocity, final velocity, acceleration due to gravity, the ball falls, 2.5m.

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