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A tennis ball is thrown from a 25 m tall building with a zero initial velocity. At the same moment, another ball is thrown from the ground vertically upward with an initial velocity of 17 m/s. At which height will the two balls meet?

Respuesta :

Answer:

The two balls meet in 1.47 sec.

Explanation:

Given that,

Height = 25 m

Initial velocity of ball= 0

Initial velocity of another ball = 17 m/s

We need to calculate the ball

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2+h[/tex]

Where, u = initial velocity

h = height

g = acceleration due to gravity

Put the value in the equation

For first ball

[tex]s_{1}=0-\dfrac{1}{2}gt^2+25[/tex]....(I)

For second ball

[tex]s_{2}=17t-\dfrac{1}{2}gt^2+0[/tex]....(II)

From equation (I) and (II)

[tex]-\dfrac{1}{2}gt^2+25=17t-\dfrac{1}{2}gt^2+0[/tex]

[tex]t=\dfrac{25}{17}[/tex]

[tex]t=1.47\ sec[/tex]

Hence, The two balls meet in 1.47 sec.

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