Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

An athlete jumps a horizontal distance of $7.18$m. He was airborne for $2.29$ seconds. The acceleration due to gravity is taken as $9.81$ms$^{-2}$. Assume that air resistance is negligible. Calculate the take-off speed in ms$^{-1}$

share|cite|improve this question
What have you tried? How are you attempting to solve the problem? Where are you getting stuck? – Matthew Conroy Mar 26 '12 at 3:00
What do you know about this kind of problem? What facts and equations have you seen that deal with this kind of situation? – Gerry Myerson Mar 26 '12 at 5:37
up vote 5 down vote accepted

The horizontal component $v_1$ of the velocity is an unchanging $\dfrac{7.18}{2.29}$ metres per second.

Maximum height is reached after $\dfrac{2.29}{2}$ seconds. If $v_2$ is the initial vertical component of the velocity, then the vertical component of the velocity, after $t$ seconds, is $v_2-9.81t$ (for $t\le 2.29$). This vertical component of the velocity reaches $0$ at time $\frac{2.29}{2}$. It follows that $$v_2-9.81\frac{2.29}{2}=0.$$ Now we know $v_1$ and $v_2$. The initial speed is $\sqrt{v_1^2+v_2^2}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.