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Chapter 3: Q45PE (page 124)

In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77m. What was the initial speed of the shot if he released it at a height of2.10m and 38.0掳threw it at an angle of above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45掳 when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus,38掳 will give a longer range than45掳 in the shot put.)

Short Answer

Expert verified

The initial speed is15m/s

Step by step solution

01

Stating given data

The shotput is released at the height of24.77 meters. The angle of the release of the ball is 38掳

cos胃=qhcos胃=VixViVix=Vicos38

The initial velocity in the y direction will be

02

Calculating angle at which ball should be thrown

The velocity in the x frame will be constant in all time. Considering the initial velocity

V炉=xtVicos胃=24.77tt=24.77Vicos38

The time taken is used from the above equation.

Putting the values into the equation of motion

role="math" localid="1668677707679" 鈭�Y=Viyt+12ayt2-2.1=(Visin胃)t+12(-9.8)t2putthevalueoft-2.1=(Visin38)24.77Vicos38-4.924.7Vicos382-2.1=19.4-4.924.77Vicos382

-2.1=19.4-4.9988Vi2-2.1=19.4-4840Vi2Vi2=-4840-21.5=225Vi=15m/s

The initial speed is 15m/s

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