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A ball thrown horizontally with speed vi = 23.0 m/s travels a horizontal distance of d = 59.0 m before hitting the ground. from what height h was the ball thrown?

Respuesta :

Kalahira
h = 32.2 m
       
Assuming no air resistance, you can determine how long the ball traveled by dividing the distance traveled by the velocity. So t = 59.0 m / 23 m/s = 2.565217 seconds. Once you know how much time was taken, you can then use the formula d = 1/2AT^2 to determine how far the ball will have dropped. Using A = 9.8m/s^2 and T = 2.565217 seconds we get h = 1/2 * 9.8 m/s^2 * (2.565217 s)^2 = 4.9 m/s^2 * (6.58 s^2) = 32.242 m Rounded to the nearest tenth, you have 32.2 m
okpalawalter8

We have that the height h from where the ball was thrown is

[tex]S=32.2382m[/tex]

From the Question we are told that

Speed [tex]v= 23.0 m/s[/tex]

Horizontal distance of [tex]d = 59.0 m[/tex]

Generally the equation for Time  is mathematically given as

[tex]T=\frac{d}{v}[/tex]

[tex]t=59/23\\\\t=2.565s[/tex]

Therefore

[tex]S=ut+1/2at^2\\\\S=1/2(9.8)(2.565)^2[/tex]

[tex]S=32.2382m[/tex]

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