If a projectile is shot at an angle of $pi/6$ and at a velocity of $24 m/s$ , when will it reach its maximum height??

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Answer 1 · 2016-06-16T20:28:44

$t~~1.223 " seconds"$ to 2 decimal places

Note that $9.81$ for g is an approximate figure so the solution is an approximation to close limits.

Tony B
Let time in seconds be $t$
Let vertical height by $h$

Vertical height at any time is

$h=[24sin(pi/6)xxt] - [1/2xx9.81xxt^2]$

The $ul("rate of change of h")$ will be 0 at maximum height

$=> (dh)/(dt)~~ 24sin(pi/6)-9.81t$

But at maximum height $(dy)/(dt)=0$ giving

$9.81t~~24sin(pi/6)$

$=> t=24/9.81sin(pi/6)$

But $sim(pi/6)=1/2$

$=>t~~12/9.81$

$t~~1.223 " seconds"$ to 2 decimal places

If a projectile is shot at an angle of pi/6pi/6 and at a velocity of 24 m/s24 m/s, when will it reach its maximum height??