A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?
A) 43 m/s
B) 47 m/s
C) 39 m/s
D) 36 m/s
E) 14 m/s

Question

Quizplus · Accepted Answer

Using kinematic equations, we can solve for the initial horizontal velocity ($v_{0x}$) of the ball:
$h = \frac{1}{2}at^{2}$
$t = \sqrt{\frac{2h}{g}}$
$R = v_{0x}t$
$v_{0x} = \frac{R}{t}$
where $h =$ height of building $= 0.10\ 	ext{km} = 100\ 	ext{m}$, $R =$ horizontal distance from point of release to point of impact $= 65\ 	ext{m}$, and $g =$ acceleration due to gravity $= 9.81\ 	ext{m/s}^{2}$.

Plugging in the values, we get:
$t = \sqrt{\frac{2(100\ 	ext{m})}{9.81\ 	ext{m/s}^{2}}} = 4.52\ 	ext{s}$
$v_{0x} = \frac{65\ 	ext{m}}{4.52\ 	ext{s}} = 14.4\ 	ext{m/s}$

Now, we can use the horizontal and vertical components of velocity to solve for the magnitude of the velocity just before impact:
$v_{x} = v_{0x} = 14.4\ 	ext{m/s}$
$v_{y}^{2} = v_{0y}^{2} + 2gh$
$v_{y} = \sqrt{2gh}$
where $h =$ height of building $= 100\ 	ext{m}$.

Plugging in the values, we get:
$v_{y} = \sqrt{2(9.81\ 	ext{m/s}^{2})(100\ 	ext{m})} = 44.3\ 	ext{m/s}$

Using the Pythagorean theorem, we get:
$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(14.4\ 	ext{m/s})^{2} + (44.3\ 	ext{m/s})^{2}} = 47.0\ 	ext{m/s}$

Therefore, the answer is (B) 47 m/s.

A Ball Is Thrown Horizontally from the Top of a Building