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A student standing on a knoll throws a snowball horizontally 4.5 meters above the level ground toward a smokestack 15 meters away. The snowball hits the smokestack 0.65 second after being released. [Neglect air resistance.]
Approximately how far above the level ground does the snowball hit the smokestack?

Respuesta :

Answer:

2.4 m

Explanation:

Consider the motion along the vertical direction

[tex]y_{o}[/tex] = initial position of ball above the ground = 4.5 m

[tex]t[/tex] = time taken by the ball to hit the smokestack = 0.65 s

[tex]v_{oy}[/tex] = initial velocity of the ball along vertical direction

[tex]a_{y}[/tex] = acceleration due to gravity = - 9.8 m/s²

[tex]y[/tex] = position of ball at the time of hitting the smokestack

Using the kinematics equation

[tex]y = y_{o} + v_{oy} t + (0.5) a_{y} t^{2}[/tex]

inserting the above values

[tex]y = 4.5 + (0) (0.65) + (0.5) (- 9.8) (0.65)^{2} \\y = 2.4 m[/tex]

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