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The speed of the toy when it hits the ground is 2.97 m/s.
The given parameters;
The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.
P.E = K.E
[tex]mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}[/tex]
Substitute the given values and solve the speed;
[tex]v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s[/tex]
Thus, the speed of the toy when it hits the ground is 2.97 m/s.
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