Question

As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 17.3 m/s. in the process, he moves his hand through a distance of 1.43 m. the acceleration of gravity is 9.8 m/s 2 . if the ball has a mass of 0.196 kg, find the force he exerts on the ball to give it this upward speed. answer in units of n.

Answer 1

While still in his hand, for the ball to attain a speed of 17.3 m/s over 1.43 m would require that the throw takes

$\dfrac{1.43\,\mathrm m}{17.3\,\frac{\mathrm m}{\mathrm s}}=0.0827\,\mathrm s$

Then the (presumably constant) acceleration the pitcher applies to the ball is

$a=\dfrac{17.3\,\frac{\mathrm m}{\mathrm s}}{0.0827\,\mathrm s}=209\,\dfrac{\mathrm m}{\mathrm s^2}$

The total acceleration acting on the ball is then

$209\,\dfrac{\mathrm m}{\mathrm s^2}-9.8\,\dfrac{\mathrm m}{\mathrm s^2}=199\,\dfrac{\mathrm m}{\mathrm s^2}$

With a mass of $0.196\,\mathrm{kg}$, the force applied to the ball would be

$F=(0.196\,\mathrm{kg})\left(199\,\dfrac{\mathrm m}{\mathrm s^2}\right)=39\,\dfrac{\mathrm{kg}\cdot\mathrm m}{\mathrm s^2}=39\,\mathrm N$

Answer 2

The force exerted on the ball to give it this speed is 18.6 N.

The given parameters;

• velocity of the ball, u = 17.3 m/s

• distance moved by the ball, h = 1.43 m

• mass of the ball, m = 0.196 kg

The acceleration of the ball is calculated as;

$v^2 = u^2 + 2as\\\\0 = 17.3^2 + (2\times - 1.43)a\\\\2.86a = 299.29\\\\a = \frac{299.29}{2.86} \\\\a = 104.65 \ m/s^2$

The force exerted on the ball to give it this speed is calculated as;

$F = m(a -g)\\\\F = 0.196(104.65 - 9.8)\\\\F = 18.6 \ N$

Thus, the force exerted on the ball to give it this speed is 18.6 N.

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