Applied physics

Classical Mechanics Level 3

A golfer strikes a ball from a tee of negligible height. 4.5 4.5 seconds later, the ball lands in a bunker 225 225 meters away. At what initial speed, in m / s m/s , was the ball hit?

Details:

• g = 10 m / s 2 g=10 m/s^2

• The ground is perfectly level throughout.

• Ignore wind and air resistance.


The answer is 54.83.

Alex Li by Alex Li , 22, USA

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1 solution

Alex Li
Apr 13, 2015

The horizontal component of its speed is 225 4.5 = 50 m / s \frac{225}{4.5}=50 m/s , while the vertical component is 4.5 × g 2 = 4.5 × 5 = 22.5 m / s \frac{4.5\times g}{2}=4.5\times5=22.5 m/s . Therefore, the overall speed is 5 0 2 + 22. 5 2 = 54.83 m / s \sqrt{50^2+22.5^2}=\boxed{54.83 m/s}

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