If a projectile is shot at an angle of $\frac{π}{4}$ $pi/4$ and at a velocity of $15 \frac{m}{s}$ $15 m/s$ , when will it reach its maximum height??

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If a projectile is shot at an angle of $\frac{π}{4}$ $pi/4$ and at a velocity of $15 \frac{m}{s}$ $15 m/s$ , when will it reach its maximum height??

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Answer 1 · 2016-02-28T02:57:30

2.12s

Explanation:

Velocity of projection $, v = 15 m s^{- 1}$ $,v=15ms^-1$
Angle of projection $, α = \frac{π}{4}$ $,alpha=pi/4$
Vertical component of vel. of projection $, v sin α = 15 \cdot sin (\frac{π}{4}) = \frac{15}{\sqrt{2}} m s^{- 1}$ $,vsinalpha=15*sin(pi/4)=15/sqrt2ms^-1$
Applying equn for vertical motion under gravity,
at maximum height (H) the final vertical vel. being ZERO, we can write
$0^{2} = {(v sin α)}^{2} - 2 \cdot g \cdot H$ $0^2=(vsinalpha)^2-2*g*H$
$\Rightarrow H = \frac{{(v sin α)}^{2}}{2 g} = \frac{15^{2}}{2^{2} \cdot 10} = 5.625 m$ $=>H=(vsinalpha)^2/(2g)=15^2/(2^2*10) =5.625m$ , [taking g = 10 $m s^{- 2}$ $ms^-2$ ]
If time of reaching Maximum height be T then
$0 = v sin α \cdot T - \frac{1}{2} g T^{2}$ $0=vsinalpha*T-1/2gT^2$
$\Rightarrow T = \frac{2 v sin α}{g} = 2 \cdot \frac{15}{10 \sqrt{2}} = 2.12 s$ $=>T=(2vsinalpha)/g=2*15/(10sqrt2)=2.12s$

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If a projectile is shot at an angle of $\frac{π}{4}$ $pi/4$ and at a velocity of $15 \frac{m}{s}$ $15 m/s$ , when will it reach its maximum height??

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Explanation:

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If a projectile is shot at an angle of $\frac{π}{4}$ $pi/4$ and at a velocity of $15 \frac{m}{s}$ $15 m/s$ , when will it reach its maximum height??