If a projectile is shot at an angle of $\frac{5 π}{12}$ $(5pi)/12$ and at a velocity of $19 \frac{m}{s}$ $19 m/s$ , when will it reach its maximum height??

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If a projectile is shot at an angle of $\frac{5 π}{12}$ $(5pi)/12$ and at a velocity of $19 \frac{m}{s}$ $19 m/s$ , when will it reach its maximum height??

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Answer 1 · 2017-04-01T11:41:57

The projectile will reach its maximum height when the velocity equals zero.

Explanation:

At an angle of $(\frac{5 π}{12})$ $((5pi)/12)$ , the initial velocity forms the hypotenuse of a right-angled triangle with angle $(\frac{5 π}{12})$ $((5pi)/12)$ at the base.
The maximum height of the shot will be the vertical component $(y)$ $(y)$ , of the initial velocity vector of $19$ $19$ m/s, and will occur when the final velocity equals zero.
The initial velocity of the vertical component is $V o y = 19 sin (\frac{5 π}{12}) = 0.434$ $Voy = 19sin((5pi)/12) = 0.434$ m/s.
Using $V = V o + a t$ $V = Vo + at$ we can rearrange to find $t$ $t$ :

$t = \frac{V - V o}{a}$ $t = (V - Vo)/a$

$= \frac{0 - 0.434}{- 9.8} = 0.04 s$ $= (0 - 0.434)/- 9.8 = 0.04 s$

Assuming that the force due to gravity is in the downward direction, therefore, the upward direction has a negative sign.

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If a projectile is shot at an angle of $\frac{5 π}{12}$ $(5pi)/12$ and at a velocity of $19 \frac{m}{s}$ $19 m/s$ , when will it reach its maximum height??

1 Answer

Explanation:

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If a projectile is shot at an angle of $\frac{5 π}{12}$ $(5pi)/12$ and at a velocity of $19 \frac{m}{s}$ $19 m/s$ , when will it reach its maximum height??