# My first classical note

A student sits atop a platform a distance h above the ground. He throws a large object horizontally with a speed $$u$$. A wind blowing parallel to the ground gives the object a constant horizontal acceleration with magnitude a. This results in the object reaching the ground directly under the student. Determine the height $$h$$ in terms of $$u$$, $$a$$ and $$g$$. Ignore the effect of air resistance on the vertical motion.

Note by Department 8
2 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

• bulleted
• list

1. numbered
2. list

1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Since there is no initial velocity, $$h=\dfrac{1}{2}gt^2$$(1).

Since the object reaches directly under the student, $$s=0=ut+\dfrac{1}{2}(-a)t^2$$(acceleration is negative as it is in the opposite direction) or $$0=u-\dfrac{1}{2}at$$ or $$t=\dfrac{2u}{a}$$(2).

Substituting the value of (2) into (1) we get $$h=\dfrac{2gu^2}{a^2}$$.

- 2 years, 3 months ago