# Chapter 3 - Problems - Page 111: 16

The speed of the baseball must be 91.5 mph

#### Work Step by Step

Let $t$ be the time it takes for the ball to reach home plate. In general: $t = \frac{d}{v}$ Let $v_b$ be the required speed of the baseball: $t = \frac{43.0~ft}{65.0~mph} = \frac{60.5~ft}{v_b}$ $v_b = (\frac{60.5~ft}{43.0~ft})~(65.0~mph)$ $v_b = (1.407)~(65.0~mph)$ $v_b = 91.5~mph$ The speed of the baseball must be 91.5 mph

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