Question

A shot putter threw an 8 lb 13 oz shot 59 ft, 6 in. Assuming the shot putter launched the shot at a 50° angle to the horizontal from 6.5 ft above the ground, what was the shot's initial speed? The acceleration due to gravity is 32 ft /seca. = The shot's initial speed was vo ft/sec. (Do not round until the final answer. Then round to the nearest tenth as needed.)

Answer 1

Consider the horizontal and vertical motion of the shot separately.

Note that there is no influence of gravity on the horizontal motion of the shot. So the distance travelled horizontally by the shot is purely due to the initial horizontal velocity of the shot.

Total horizotal distance travelled is 59 feet 6 inches. In other words, 59.5 feets.

Now, let the total time the shot was in the air be t.

So: the distance travelled can be expressed as a product of horizontal component of initial velocity and the total travel time. (horizotal component of initial velocity is $v_0\,cos50$ )

So

$59.5=v_0\,cos50\times t$......(1)

Now, consider the vertical motion of the shot: As it was influenced by the gravity, the total vertical distance can be expressed using the equation of motion:

$h=\left (v_o\,sin50 \times t \right )-\frac{1}{2}gt^2$.......($v_o\,sin50$ is the vertical component of initial velocity)

Note that total vertical distance travelled was 6.5 ft downwards and g is given as 32 ft.

So:

$-6.5=\left (v_o\,sin50 \times t \right )-\frac{1}{2}32t^2$

or $-6.5=\left (v_o\,sin50 \times t \right )-16\,t^2$ ......(2)

........

Now, consider equations (1) and (2).

There are 2 equations in two variables. So they can be solved to find $v_0$ and t. (You may use a calculator to solve it directly)

So consider the equation (1)

rearranging the equatin (1), $v_0\,\times t=\frac{59.5}{cos50}$

Substituting this value in equation (2) gives:

$-6.5=\frac{59.5}{cos50}sin50-16\,t^2$

$-6.5=59.5\,tan50-16\,t^2$

$16\,t^2=59.5\,tan50+6.5$

$t^2=\frac{59.5\,tan50+6.5}{16}$

$t=2.1995644$....(3)

Now, to find $v_0$ , substitute (3) directly to equation (1):

$59.5=v_0\,cos50\times 2.1995644$

i.e. on rearranging;  $v_0=\frac{59.5}{cos50\times 2.1995644}=42.08\,ft/sec$

Rounding to nearest tenth, the initial speed is 42.1 ft/sec

Therefore the shot's initial speed is 42.1 ft/sec

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