Waste less time on Facebook — follow Brilliant.
×

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 Lakshya Sinha
1 year, 8 months ago

No vote yet
1 vote

  Easy Math Editor

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

[example link](https://brilliant.org)example link
> 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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 \(h=\dfrac{2gu^2}{a^2}\).

Brilliant Member - 1 year, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...