Question
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

a) $3.50 \textrm{ s}$

b) $28.6 \textrm{ m/s}$

c) $-34.3 \textrm{ m/s}$

d) $44.6 \textrm{ m/s, } 50.2^\circ \textrm{ below the + x-axis}$

Solution Video

# OpenStax College Physics Solution, Chapter 3, Problem 27 (Problems & Exercises) (7:49)

## Calculator Screenshots  Submitted by Helen on Tue, 02/11/2020 - 08:15

Hello! My version of the first equation you use is $y=y0+v0t-1/2at^2$ , but you always show it with as $y=y0+v0t+1/2a^2$ with a plus sign between $v0t$ and $1/2a^2$ instead of a minus. Why is this?

Submitted by ShaunDychko on Tue, 02/11/2020 - 09:50

Hi Helen, good question. I think, but correct me if I'm wrong, what you meant to say was with a $g$, not an $a$, as $y = y_o + v_ot - 1/2 g t^2$. Would you agree? The version I'm showing is more generic, and $y$ can represent any position, not just the vertical position that $y$ is traditionally associated with. In your version, however, $y$ specifically refers to the vertical direction, assumes that upwards is the positive direction (and consequently down is negative), and that the acceleration takes place near the surface of Earth, in which case $a$ becomes $-g$. The short answer is that your equation is the same as mine, except with $-g$ substituted in place of $a$.
All the best,
Shaun