If a projectile is shot at an angle of $\frac{π}{4}$ $pi/4$ and at a velocity of $12 \frac{m}{s}$ $12 m/s$ , when will it reach its maximum height??

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Answer 1 · 2017-01-15T11:10:37

The maximum height is $= 3.67 m$ $=3.67m$

The vertical motion is ${↑ ⏐ ⏐}^{+}$ $uarr^+$

$u = 12 sin (\frac{π}{4}) m s^{- 1}$ $u=12sin(pi/4)ms^-1$

$v = 0 m s^{- 1}$ $v=0ms^-1$

$a = - g = - 9.8 m s^{- 2}$ $a=-g=-9.8ms^-2$

We use the equation

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

$0 = 144 {sin}^{2} (\frac{π}{4}) - 2 \cdot 9.8 \cdot h$ $0=144sin^2(pi/4)-2*9.8*h$

$h = \frac{144 \cdot {sin}^{2} (\frac{π}{4})}{2 \cdot 9.8} = 3.67 m$ $h=(144*sin^2(pi/4))/(2*9.8)=3.67m$

If a projectile is shot at an angle of pi/4π4pi/4 and at a velocity of 12 m/s12ms12 m/s, when will it reach its maximum height??