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You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 30.5 ∘ from the ground with an initial speed of 20.5 m / s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.

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opudodennis

Answer:

5.39 m

Explanation:

From kinematics

[tex]s=u_x t[/tex] where t is time, s is distance and [tex]u_x[/tex] is initial speed in x direction

[tex]u_x= 20.5cos 30.5=17.6634 m/s[/tex]

[tex]u_y= 20.5 sin 30.5=10.40454 m/s[/tex]

Now 17.5=17.6634 t

[tex]t=\frac {17.5}{17.6634}[/tex] =0.99 s

Using the equation of kinematics

[tex]h=u_y t- 0.5gt^{2}[/tex]

[tex]h=10.40454\times 0.99^{2}-0.5\times 9.81\times 0.99^{2}[/tex]

h=5.39 m

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