Question

Suppose Sammy Sosa hits a home run which travels 361. ft (110. m). Leaving the bat at 50 degrees above the horizontal, how high did it travel?

Answer 1

Answer:

The horizontal distance of Sosa is 276.526 ft or 84.28 meter.

Explanation:

As shown in the figure, let point O is the starting point of Sosa. She travels 361 ft at an angle 50 degree with the horizontal.

sin 50 = $\frac{OM}{OP}$

0.7660 = h / 361

h = 276.526 ft

h = 84.28 meter

The horizontal distance of Sosa is 276.526 ft or 84.28 meter.